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][1]. 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.

  [1]: http://en.wikipedia.org/wiki/Logarithmically_concave_measure