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Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (I will give a more complete proof of the case with general $f$ further below, once I have written it out). I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball in Wiener space.

Here's a proof of the general case, based on the following lemma which generalizes Lemma 1 above (I have a proof of this, and will update the answer once I have written it out).

Lemma 2: Suppose that $X$ satisfies an SDE of the form $dX=g(X_t,t)dt+dW$, $X_0=0$, where $g\colon\mathbb{R}^n\times[0,1]\to\mathbb{R}^n$. is (jointly) continuous and $g(x,t)$ has (jointly) continuous first and second order derivatives with respect to $x$. Then, for any continuous path $\phi\colon[0,1]\to\mathbb{R}^d$, $$ \limsup_{\epsilon\to0}\frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le\exp\left(\frac12\sup_t\lVert Df(\phi(t),t)\rVert+c\int_0^1\lVert D^2g(\phi(t),t)\rVert dt\right)$$ where $c$ is some constant (depending on $n$, but not on $g$ or the path $\phi$).

Let's use this to prove the result in the question. For any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$, define $Y$ by $Y_0=0$ and $dY=f(Y+\gamma)dt+dW$. We see that $X-\gamma$ satisfies the same SDE as $Y$, but with $W-\gamma$ in place of $W$. Applying a Girsanov transformation as above, $$ \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\frac{\mathbb{P}(X-\gamma\in B_\epsilon+\phi-\gamma)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dW_t+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dY_t-\dot\gamma f(Y_t+\gamma)dt+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]. $$ Taking limits as $\epsilon$ goes to 0 and applying Lemma 2, $$ \limsup_{\epsilon\to0} \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le K\exp\left(\int_0^1\left(\frac12\dot\gamma^2-\dot\gamma\dot\phi-\dot\gamma f(\phi)\right)dt\right). $$ As above, we use integration by parts to take the limit, and $\dot\phi$ is the derivative in the sense of distributions. The term $K$ is the right hand side of the inequality in Lemma 2, evaluated with $g(x,t)=f(x+\gamma_t,t)$ evaluated along the path $\phi-\gamma$. This is the same as the right hand side of the inequality evaluated with $g(x,t)=f(x)$ along the path $\phi$, so is independent of $\gamma$. Letting $\gamma$ be smooth approximations to $\phi$, the right hand side of the above inequality can be made arbitrarily close to 0, giving the required result.

Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (The proof with general $f$ is a bit more involved, and I give this further below). I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball in Wiener space.

Here's a proof of the general case, based on the following lemma which generalizes Lemma 1 above (a proof is given further below).

Lemma 2: Suppose that $X$ satisfies an SDE of the form $dX=g(X_t,t)dt+dW$, $X_0=0$, where $g\colon\mathbb{R}^n\times[0,1]\to\mathbb{R}^n$. is (jointly) continuous and $g(x,t)$ has (jointly) continuous first and second order derivatives with respect to $x$. Then, for any continuous path $\phi\colon[0,1]\to\mathbb{R}^d$, $$ \limsup_{\epsilon\to0}\frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le K$$ where $K$ is some increasing function of $\lVert Dg(\phi,\cdot)\rVert$ and $\lVert D^2g(\phi,\cdot)\rVert$.

Let's use this to prove the result in the question. For any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$, define $Y$ by $Y_0=0$ and $dY=f(Y+\gamma)dt+dW$. We see that $X-\gamma$ satisfies the same SDE as $Y$, but with $W-\gamma$ in place of $W$. Applying a Girsanov transformation as above, $$ \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\frac{\mathbb{P}(X-\gamma\in B_\epsilon+\phi-\gamma)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dW_t+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dY_t-\dot\gamma f(Y_t+\gamma)dt+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]. $$ Taking limits as $\epsilon$ goes to 0 and applying Lemma 2, $$ \limsup_{\epsilon\to0} \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le K\exp\left(\int_0^1\left(\frac12\dot\gamma^2-\dot\gamma\dot\phi-\dot\gamma f(\phi)\right)dt\right). $$ As above, we use integration by parts to take the limit, and $\dot\phi$ is the derivative in the sense of distributions. The term $K$ is the right hand side of the inequality in Lemma 2, evaluated with $g(x,t)=f(x+\gamma_t,t)$ evaluated along the path $\phi-\gamma$. This is the same as the right hand side of the inequality evaluated with $g(x,t)=f(x)$ along the path $\phi$, so is independent of $\gamma$. Letting $\gamma$ be smooth approximations to $\phi$, the right hand side of the above inequality can be made arbitrarily close to 0, giving the required result.

Proof of Lemma 2:

Note that the SDE can be expressed as $W_t=X_t-\int_0^tg(X_s,s)ds$. Let us define the path $\psi_t=\phi_t-\int_0^tg(\phi_s,s)ds$. Expanding $g(X_s,s)$ as a power series to second order in $X-\phi$ gives $$ W_t-\psi_t=X_t-\phi_t-\int_0^t\left(A_s(X_s-\phi_s)+u_s\right)ds, $$ where $A_s=Dg(\phi_s,s)$ and $u_s$ is of second order in $X_s-\phi_s$. Choosing a fixed $L > \lVert D^2g(\phi_t,t)\rVert/2$ then $\lVert u_s\rVert\le L\lVert X_s-\phi_s\rVert^2\le L\epsilon^2$ for $X\in B_\epsilon+\phi$ and small enough $\epsilon$. Now, define $S$ to be set of all paths $\omega$ of the form $$ \omega_t=y_t-\int_0^t(A_sy_s+v_s)ds $$ for some $y\in B_\epsilon$ and $v\in B_{L\epsilon^2}$. From the above, we have $W\in S+\psi$ whenever $X\in B_\epsilon+\phi$. As $S$ is convex and symmetric, using log-concavity as in the proof of Lemma 1, $\mathbb{P}(W\in S+\psi)\le\mathbb{P}(W\in S)$.

Now, define a process $Y$ by $$ dY_t=dW_t+AY_t,\ \ Y_0=0. $$ Suppose that $\omega\in S$, $y\in B_\epsilon$ and $v\in B_{L\epsilon^2}$ are as above and $W_t=\omega_t$ then, we have $$ d\lVert Y_t-y_t\rVert/dt\le\lVert A(Y_t-y_t)-v_t\rVert\le a\lVert Y_t-y_t\rVert+L\epsilon^2 $$ where $a=\sup_t\lVert A_t\rVert$. From this we can bound $Y_t-y_t$ by $L\epsilon^2a^{-1}(e^{at}-1)$, which I will write as $M\epsilon^2$. So, $Y\in B_{\epsilon+M\epsilon^2}$, and we have $\mathbb{P}(W\in S)\le\mathbb{P}(Y\in B_{\epsilon+M\epsilon^2})$.

Next, define the process $$ dZ = dW - aZdt,\ \ Z_0=0. $$ This has drift $-a\lVert Z\rVert$ in the radial direction, whereas $Y$ has radial drift bounded below by $-a\lVert Y\rVert$. By comparing these processes we have $\mathbb{P}(Y\in B_{\epsilon+M\epsilon^2})\le\mathbb{P}(Z\in B_{\epsilon+M\epsilon^2})$.

Now, applying the Onsager--Machlup functional to the simple case with $f(x)=-ax$ and $\phi=0$ gives $$ \lim_{\epsilon\to0}\frac{\mathbb{P}(Z\in B_{\epsilon+M\epsilon^2})}{\mathbb{P}(W\in B_{\epsilon+M\epsilon^2})}=\exp(a). $$ We can also use the standard result that $\mathbb{P}(W\in B_\epsilon)\sim\exp(-c/\epsilon)$ in the limit $\epsilon\to0$ (where $c$ depends only on the dimension $n$) to get $$ \lim_{\epsilon\to0}\frac{\mathbb{P}(W\in B_{\epsilon+M\epsilon^2})}{\mathbb{P}(W\in B_{\epsilon})}=\exp(cM). $$

Putting the above results together gives the following sequence of inequalities, which should be understood in the asymptotic limit $\epsilon\to0$. $$ \begin{align} \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)} &\le \frac{\mathbb{P}(W\in S+\psi)}{\mathbb{P}(W\in B_\epsilon)} \le \frac{\mathbb{P}(W\in S)}{\mathbb{P}(W\in B_\epsilon)}\\ &\le \frac{\mathbb{P}(Y\in B_{\epsilon+M\epsilon^2})}{\mathbb{P}(W\in B_\epsilon)} \le \frac{\mathbb{P}(Z\in B_{\epsilon+M\epsilon^2})}{\mathbb{P}(W\in B_\epsilon)}\\ &\le \exp(a)\frac{\mathbb{P}(W\in B_{\epsilon+M\epsilon^2})}{\mathbb{P}(W\in B_\epsilon)} \le\exp(a+cM). \end{align} $$ QED

Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (I will give a more complete proof of the case with general $f$ further below, once I have written it out). I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball in Wiener space.

Lemma 1: For any continuous $\gamma\colon[0,1]\to\mathbb{R}^n$ and $\epsilon > 0$ $$ \mathbb{P}\left(W\in B_\epsilon+\gamma\right)\le\mathbb{P}\left(W\in B_\epsilon\right). $$

Proof: As the Wiener measure is Gaussian, it is log-concave. As $B_\epsilon$ is convex then its indicator function is log-concave and, as the convolution of log-concave functions is log-concave, this implies that $\gamma\mapsto\mathbb{P}(W\in B_\epsilon+\gamma)$ is log-concave. As it is also symmetric this gives $$ \mathbb{P}\left(W\in B_\epsilon\right)\ge\sqrt{\mathbb{P}\left(W\in B_\epsilon+\gamma\right)\mathbb{P}\left(W\in B_\epsilon-\gamma\right)}=\mathbb{P}\left(W\in B_\epsilon+\gamma\right). $$ QED

Choosing any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$ with $\gamma(0)=0$ then, using a Girsanov transform in the usual way, $$ \begin{align} &\mathbb{P}\left(W\in B_\epsilon+\phi\right) = \mathbb{P}\left(W-\gamma\in B_\epsilon+\phi-\gamma\right)\\ &\qquad=\mathbb{E}\left[\exp\left(-\int_0^1\dot\gamma dW-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi-\gamma\right]\mathbb{P}(W\in B_\epsilon+\phi-\gamma)\\ &\qquad\le\mathbb{E}\left[\exp\left(\int_0^1\ddot\gamma W dt-\dot\gamma(1)W(1)-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi-\gamma\right]\mathbb{P}(W\in B_\epsilon) \end{align} $$ The last inequality here is using the lemma, together with integration by parts in the exponent. So, letting $\epsilon$ go to zero and using uniform convergence in the expectation, $$ \begin{align} \limsup_{\epsilon\to0}\frac{\mathbb{P}(W\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)} &\le\exp\left(\int_0^1\ddot\gamma (\phi-\gamma) dt-\dot\gamma(1)(\phi(1)-\gamma(1))-\frac12\int_0^1\dot\gamma^2dt\right)\\ &=\exp\left(\frac12\int_0^1\dot\gamma^2dt-\int_0^1\dot\gamma\dot\phi dt\right). \end{align} $$ In the last line, the derivative $\dot\phi$ is understood in the sense of distributions. Now, if $\phi$ is not in Cameron--Martin space, then if we let $\gamma$ be smooth approximations to $\phi$ (e.g., convolve with a smooth bump function), the right hand side tends to zero giving the result.

Here's a proof of the general case, based on the following lemma which generalizes Lemma 1 above (I have a proof of this, and will update the answer once I have written it out).

Lemma 2: Suppose that $X$ satisfies an SDE of the form $dX=g(X_t,t)dt+dW$, $X_0=0$, where $g\colon\mathbb{R}^n\times[0,1]\to\mathbb{R}^n$. is (jointly) continuous and $g(x,t)$ has (jointly) continuous first and second order derivatives with respect to $x$. Then, for any continuous path $\phi\colon[0,1]\to\mathbb{R}^d$, $$ \limsup_{\epsilon\to0}\frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le\exp\left(\int_0^1\left(-\frac12\operatorname{div} g(\phi(t),t)+c\lVert D^2g(\phi(t),t)\rVert\right)dt\right)$$ where $c$ is some constant (depending on $n$, but not on $g$ or the path $\phi$).

Let's use this to prove the result in the question. For any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$, define $Y$ by $Y_0=0$ and $dY=f(Y+\gamma)dt+dW$. We see that $X-\gamma$ satisfies the same SDE as $Y$, but with $W-\gamma$ in place of $W$. Applying a Girsanov transformation as above, $$ \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\frac{\mathbb{P}(X-\gamma\in B_\epsilon+\phi-\gamma)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dW_t+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dY_t-\dot\gamma f(Y_t+\gamma)dt+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]. $$ Taking limits as $\epsilon$ goes to 0 and applying Lemma 2, $$ \limsup_{\epsilon\to0} \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le K\exp\left(\int_0^1\left(\frac12\dot\gamma^2-\dot\gamma\dot\phi-\dot\gamma f(\phi)\right)dt\right). $$ As above, we use integration by parts to take the limit, and $\dot\phi$ is the derivative in the sense of distributions. The term $K$ is the right hand side of the inequality in Lemma 2, evaluated with $g(x,t)=f(x+\gamma_t,t)$ evaluated along the path $\phi-\gamma$. This is the same as the right hand side of the inequality evaluated with $g(x,t)=f(x)$ along the path $\phi$, so is independent of $\gamma$. Letting $\gamma$ be smooth approximations to $\phi$, the right hand side of the above inequality can be made arbitrarily close to 0, giving the required result.

Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (I will give a more complete proof of the case with general $f$ further below, once I have written it out). I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball in Wiener space.

Lemma 1: For any continuous $\gamma\colon[0,1]\to\mathbb{R}^n$ and $\epsilon > 0$ $$ \mathbb{P}\left(W\in B_\epsilon+\gamma\right)\le\mathbb{P}\left(W\in B_\epsilon\right). $$

Proof: As the Wiener measure is Gaussian, it is log-concave. As $B_\epsilon$ is convex then its indicator function is log-concave and, as the convolution of log-concave functions is log-concave, this implies that $\gamma\mapsto\mathbb{P}(W\in B_\epsilon+\gamma)$ is log-concave. As it is also symmetric this gives $$ \mathbb{P}\left(W\in B_\epsilon\right)\ge\sqrt{\mathbb{P}\left(W\in B_\epsilon+\gamma\right)\mathbb{P}\left(W\in B_\epsilon-\gamma\right)}=\mathbb{P}\left(W\in B_\epsilon+\gamma\right). $$ QED

Choosing any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$ with $\gamma(0)=0$ then, using a Girsanov transform in the usual way, $$ \begin{align} &\mathbb{P}\left(W\in B_\epsilon+\phi\right) = \mathbb{P}\left(W-\gamma\in B_\epsilon+\phi-\gamma\right)\\ &\qquad=\mathbb{E}\left[\exp\left(-\int_0^1\dot\gamma dW-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi-\gamma\right]\mathbb{P}(W\in B_\epsilon+\phi-\gamma)\\ &\qquad\le\mathbb{E}\left[\exp\left(\int_0^1\ddot\gamma W dt-\dot\gamma(1)W(1)-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi-\gamma\right]\mathbb{P}(W\in B_\epsilon) \end{align} $$ The last inequality here is using the lemma, together with integration by parts in the exponent. So, letting $\epsilon$ go to zero and using uniform convergence in the expectation, $$ \begin{align} \limsup_{\epsilon\to0}\frac{\mathbb{P}(W\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)} &\le\exp\left(\int_0^1\ddot\gamma (\phi-\gamma) dt-\dot\gamma(1)(\phi(1)-\gamma(1))-\frac12\int_0^1\dot\gamma^2dt\right)\\ &=\exp\left(\frac12\int_0^1\dot\gamma^2dt-\int_0^1\dot\gamma\dot\phi dt\right). \end{align} $$ In the last line, the derivative $\dot\phi$ is understood in the sense of distributions. Now, if $\phi$ is not in Cameron--Martin space, then if we let $\gamma$ be smooth approximations to $\phi$ (e.g., convolve with a smooth bump function), the right hand side tends to zero giving the result.

Here's a proof of the general case, based on the following lemma which generalizes Lemma 1 above (I have a proof of this, and will update the answer once I have written it out).

Lemma 2: Suppose that $X$ satisfies an SDE of the form $dX=g(X_t,t)dt+dW$, $X_0=0$, where $g\colon\mathbb{R}^n\times[0,1]\to\mathbb{R}^n$. is (jointly) continuous and $g(x,t)$ has (jointly) continuous first and second order derivatives with respect to $x$. Then, for any continuous path $\phi\colon[0,1]\to\mathbb{R}^d$, $$ \limsup_{\epsilon\to0}\frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le\exp\left(\frac12\sup_t\lVert Df(\phi(t),t)\rVert+c\int_0^1\lVert D^2g(\phi(t),t)\rVert dt\right)$$ where $c$ is some constant (depending on $n$, but not on $g$ or the path $\phi$).

Let's use this to prove the result in the question. For any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$, define $Y$ by $Y_0=0$ and $dY=f(Y+\gamma)dt+dW$. We see that $X-\gamma$ satisfies the same SDE as $Y$, but with $W-\gamma$ in place of $W$. Applying a Girsanov transformation as above, $$ \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\frac{\mathbb{P}(X-\gamma\in B_\epsilon+\phi-\gamma)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dW_t+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dY_t-\dot\gamma f(Y_t+\gamma)dt+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]. $$ Taking limits as $\epsilon$ goes to 0 and applying Lemma 2, $$ \limsup_{\epsilon\to0} \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le K\exp\left(\int_0^1\left(\frac12\dot\gamma^2-\dot\gamma\dot\phi-\dot\gamma f(\phi)\right)dt\right). $$ As above, we use integration by parts to take the limit, and $\dot\phi$ is the derivative in the sense of distributions. The term $K$ is the right hand side of the inequality in Lemma 2, evaluated with $g(x,t)=f(x+\gamma_t,t)$ evaluated along the path $\phi-\gamma$. This is the same as the right hand side of the inequality evaluated with $g(x,t)=f(x)$ along the path $\phi$, so is independent of $\gamma$. Letting $\gamma$ be smooth approximations to $\phi$, the right hand side of the above inequality can be made arbitrarily close to 0, giving the required result.

Here's a proof of the statement for $f=0$ (so that $X=W$ is a Wiener process). As the terms involving $f$ remain well-behaved for any continuous $\phi$, I think that it should be possible to extend the usual proofs to include nonzero $f$ and arbitrary $\phi$. I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball in Wiener space.

Lemma: For any continuous $\gamma\colon[0,1]\to\mathbb{R}^n$ and $\epsilon > 0$ $$ \mathbb{P}\left(W\in B_\epsilon+\gamma\right)\le\mathbb{P}\left(W\in B_\epsilon\right). $$

Proof: As the Wiener measure is Gaussian, it is log-concave. As $B_\epsilon$ is convex then its indicator function is log-concave and, as the convolution of log-concave functions is log-concave, this implies that $\gamma\mapsto\mathbb{P}(W\in B_\epsilon+\gamma)$ is log-concave. As it is also symmetric this gives $$ \mathbb{P}\left(W\in B_\epsilon\right)\ge\sqrt{\mathbb{P}\left(W\in B_\epsilon+\gamma\right)\mathbb{P}\left(W\in B_\epsilon-\gamma\right)}=\mathbb{P}\left(W\in B_\epsilon+\gamma\right). $$ QED

Choosing any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$ with $\gamma(0)=0$ then, using a Girsanov transform in the usual way, $$ \begin{align} &\mathbb{P}\left(W\in B_\epsilon+\phi\right) = \mathbb{P}\left(W+\gamma\in B_\epsilon+\phi+\gamma\right)\\ &\qquad=\mathbb{E}\left[\exp\left(\int_0^1\dot\gamma dW-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi+\gamma\right]\mathbb{P}(W\in B_\epsilon+\phi+\gamma)\\ &\qquad\le\mathbb{E}\left[\exp\left(\dot\gamma(1)W(1)-\int_0^1\ddot\gamma W dt-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi+\gamma\right]\mathbb{P}(W\in B_\epsilon) \end{align} $$ The last inequality here is using the lemma, together with integration by parts in the exponent. So, letting $\epsilon$ go to zero and using uniform convergence in the expectation, $$ \begin{align} \limsup_{\epsilon\to0}\frac{\mathbb{P}(W\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)} &\le\exp\left(\dot\gamma(1)(\phi(1)+\gamma(1))-\int_0^1\ddot\gamma (\phi+\gamma) dt-\frac12\int_0^1\dot\gamma^2dt\right)\\ &=\exp\left(\frac12\int_0^1\dot\gamma^2dt+\int_0^1\dot\gamma\dot\phi dt\right). \end{align} $$ In the last line, the derivative $\dot\phi$ is understood in the sense of distributions. Now, if $\phi$ is not in Cameron--Martin space, then if we let $\gamma$ be smooth approximations to $-\phi$ (e.g., convolve with a smooth bump function), the right hand side tends to zero giving the result.

Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (I will give a more complete proof of the case with general $f$ further below, once I have written it out). I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball in Wiener space.

Lemma 1: For any continuous $\gamma\colon[0,1]\to\mathbb{R}^n$ and $\epsilon > 0$ $$ \mathbb{P}\left(W\in B_\epsilon+\gamma\right)\le\mathbb{P}\left(W\in B_\epsilon\right). $$

Proof: As the Wiener measure is Gaussian, it is log-concave. As $B_\epsilon$ is convex then its indicator function is log-concave and, as the convolution of log-concave functions is log-concave, this implies that $\gamma\mapsto\mathbb{P}(W\in B_\epsilon+\gamma)$ is log-concave. As it is also symmetric this gives $$ \mathbb{P}\left(W\in B_\epsilon\right)\ge\sqrt{\mathbb{P}\left(W\in B_\epsilon+\gamma\right)\mathbb{P}\left(W\in B_\epsilon-\gamma\right)}=\mathbb{P}\left(W\in B_\epsilon+\gamma\right). $$ QED

Choosing any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$ with $\gamma(0)=0$ then, using a Girsanov transform in the usual way, $$ \begin{align} &\mathbb{P}\left(W\in B_\epsilon+\phi\right) = \mathbb{P}\left(W-\gamma\in B_\epsilon+\phi-\gamma\right)\\ &\qquad=\mathbb{E}\left[\exp\left(-\int_0^1\dot\gamma dW-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi-\gamma\right]\mathbb{P}(W\in B_\epsilon+\phi-\gamma)\\ &\qquad\le\mathbb{E}\left[\exp\left(\int_0^1\ddot\gamma W dt-\dot\gamma(1)W(1)-\frac12\int_0^1\dot\gamma^2dt\right)\Bigg\vert W\in B_\epsilon+\phi-\gamma\right]\mathbb{P}(W\in B_\epsilon) \end{align} $$ The last inequality here is using the lemma, together with integration by parts in the exponent. So, letting $\epsilon$ go to zero and using uniform convergence in the expectation, $$ \begin{align} \limsup_{\epsilon\to0}\frac{\mathbb{P}(W\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)} &\le\exp\left(\int_0^1\ddot\gamma (\phi-\gamma) dt-\dot\gamma(1)(\phi(1)-\gamma(1))-\frac12\int_0^1\dot\gamma^2dt\right)\\ &=\exp\left(\frac12\int_0^1\dot\gamma^2dt-\int_0^1\dot\gamma\dot\phi dt\right). \end{align} $$ In the last line, the derivative $\dot\phi$ is understood in the sense of distributions. Now, if $\phi$ is not in Cameron--Martin space, then if we let $\gamma$ be smooth approximations to $\phi$ (e.g., convolve with a smooth bump function), the right hand side tends to zero giving the result.

Here's a proof of the general case, based on the following lemma which generalizes Lemma 1 above (I have a proof of this, and will update the answer once I have written it out).

Lemma 2: Suppose that $X$ satisfies an SDE of the form $dX=g(X_t,t)dt+dW$, $X_0=0$, where $g\colon\mathbb{R}^n\times[0,1]\to\mathbb{R}^n$. is (jointly) continuous and $g(x,t)$ has (jointly) continuous first and second order derivatives with respect to $x$. Then, for any continuous path $\phi\colon[0,1]\to\mathbb{R}^d$, $$ \limsup_{\epsilon\to0}\frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le\exp\left(\int_0^1\left(-\frac12\operatorname{div} g(\phi(t),t)+c\lVert D^2g(\phi(t),t)\rVert\right)dt\right)$$ where $c$ is some constant (depending on $n$, but not on $g$ or the path $\phi$).

Let's use this to prove the result in the question. For any smooth $\gamma\colon[0,1]\to\mathbb{R}^n$, define $Y$ by $Y_0=0$ and $dY=f(Y+\gamma)dt+dW$. We see that $X-\gamma$ satisfies the same SDE as $Y$, but with $W-\gamma$ in place of $W$. Applying a Girsanov transformation as above, $$ \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\frac{\mathbb{P}(X-\gamma\in B_\epsilon+\phi-\gamma)}{\mathbb{P}(Y\in B_\epsilon+\phi-\gamma)}=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dW_t+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]=\mathbb{E}\left[\exp\left(-\int_0^1\left(\dot\gamma dY_t-\dot\gamma f(Y_t+\gamma)dt+\frac12\dot\gamma^2dt\right)\right)\Bigg\vert Y\in B_\epsilon+\phi-\gamma\right]. $$ Taking limits as $\epsilon$ goes to 0 and applying Lemma 2, $$ \limsup_{\epsilon\to0} \frac{\mathbb{P}(X\in B_\epsilon+\phi)}{\mathbb{P}(W\in B_\epsilon)}\le K\exp\left(\int_0^1\left(\frac12\dot\gamma^2-\dot\gamma\dot\phi-\dot\gamma f(\phi)\right)dt\right). $$ As above, we use integration by parts to take the limit, and $\dot\phi$ is the derivative in the sense of distributions. The term $K$ is the right hand side of the inequality in Lemma 2, evaluated with $g(x,t)=f(x+\gamma_t,t)$ evaluated along the path $\phi-\gamma$. This is the same as the right hand side of the inequality evaluated with $g(x,t)=f(x)$ along the path $\phi$, so is independent of $\gamma$. Letting $\gamma$ be smooth approximations to $\phi$, the right hand side of the above inequality can be made arbitrarily close to 0, giving the required result.