It is not the case that the support is a singleton in general, and in fact I believe the support will be generically full. I think the full picture is a bit subtle, but let me give you some elements to explore what this support could be.

Let me first paint you the crudest picture of rough path theory. Be aware that I will use slight variations on the standard terminology. Basically, one is interested in the solutions to some controlled equation $$ \frac{\mathrm d}{\mathrm dt}X^w_t = \mu(X_t) + \sigma(X_t)\frac{\mathrm d}{\mathrm dt}w_t $$ in $\mathbb R^d$, where $w$ is a given smooth path with values in $\mathbb R^k$. $\mu$ is a vector field on $\mathbb R^d$, and $\sigma$ is a map with values in $L(\mathbb R^k,\mathbb R^d)$, that we assume to be smooth for simplicity.

Namely, a rough path lift of a smooth path is a map of the form $$ (s,t)\mapsto w_t-w_s,\Big(\int_s^t(w^i_u-w^i_s)\frac{\mathrm d}{\mathrm du}w^j_u\Big)_{i,j}\mathrm du $$ with values in $\mathbb R^d\times\mathbb R^{d\times d}$. This is called the (2-step) rough path lift of $w$. We usually write $\mathbb R^d\otimes\mathbb R^d$ for $\mathbb R^{d\times d}$ and $u\otimes v$ for $(u^iv^j)_{i,j}$, for reasons that are extremely natural once one is familiar with tensor algebras. If we write $\mathbf w$ for this pair, the result is that $\mathbf w\mapsto X^w$, which a priori does not depend on the second component, is actually continuous with respect to topologies on the space of lifts of smooth paths that are of practical interest. In some sense, they build up on the $\mathcal C^\alpha$ topology, for $1/3<\alpha<1/2$, but I don't want to go into details; it is however important that for a fixed $\alpha$, it is induced by an explicit metric, and it is stronger than the uniform topology. In particular, it admits a completion $\mathsf{RP}^\alpha([0,1],\mathbb R^d)$ that we can identify with a subset of $\mathcal C^0([0,1]^2,\mathbb R^d\times\mathbb R^d\otimes\mathbb R^d)$. This completion is the set of geometric rough paths of regularity $\alpha$, and the continuous extension of the solution map is called the rough solution map.

How does this connect to the problem at hand? Well because of the continuity of the solution map, we know that the support of $X$ when $W$ is conditioned to a tube is the closure of the image of the support of $\mathbf W$ under the same conditioning. However, this is known to contain the lift of any smooth curve with values in the tube, and in fact to be the closure of the set of such lifts. Let us describe some of these elements. If we call smooth geometric rough paths the set of elements $(a,b)$ in $\mathsf{RP}^\alpha$ which are represented by elements of $\mathcal C^\infty([0,1]^2,\mathbb R^d\times\mathbb R^d\otimes\mathbb R^d)$, then they satisfying the algebraic conditions $$\begin{aligned} a_{s,t} &= a_{s,u}+a_{u,t},\\ b_{s,t} &= b_{s,u}+b_{u,t}+a_{u,t}\otimes a_{s,u},\\ b_{s,t}+b_{s,t}^* &= a_{s,t}\otimes a_{s,t}. \end{aligned}$$ Notice that we can replace $(a,b)$ by $(a,b+\beta)$, where $\beta_{s,t}=\beta_{0,t}-\beta_{0,s}$ and $\beta_{0,t}$ is antisymmetric, and still get a smooth geometric rough path; conversely, if $(a,b)$ is a smooth geometric rough path and we define $w:t\mapsto a_{0,t}$, there exists a $\beta$ antisymmetric such that $\beta_{s,t}=\beta_{0,t}-\beta_{0,s}$ and $(a,b)=\mathbf w+(0,\beta)$, where $\mathbf w$ is defined as above. An important remark is for every $w$ and $\beta$ as above, there exist smooth curves converging to $w$ in the continuous topology such that their rough path lifts converge to $w+(0,\beta)$. This proves that the support of $X$ conditionned to $W$ staying in some open set of the uniform topology contains the rough solution of the controlled equation driven by all the smooth geometric rough paths $\mathbf w+(0,\beta)$, for $w$ in the open set.

The solution to the controlled equation driven by $\mathbf w+(0,\beta)$ is easy to describe; it is the solution to the ordinary differential equation $$ \frac{\mathrm d}{\mathrm dt}X_t = \sigma(X_t)\frac{\mathrm d}{\mathrm dt}w_t + \mu(X_t) + \frac12[\sigma,\sigma](X_t)\frac{\mathrm d}{\mathrm dt}\beta_{0,t}, $$ where $[\sigma,\sigma](x)$ is the linear endomorphism sending $u\otimes v$ to the Lie bracket at $x$ of the vector fields $y\mapsto\sigma(y)u$ and $y\mapsto\sigma(y)v$. But the image of $[\sigma,\sigma](x)$ can be much larger than that of $\sigma(x)$. Let me give an example where we can show that the support is bigger than expected.

  Choose $k=2$, $d=3$, $\mu\equiv0$, $\sigma(x)u=u^1\partial_{x^1}+u^2\partial_{x^2}+(u^1x^2-u^2x^1)\partial_{x^3}$, $f\equiv0$. Then $$ (s,t)\mapsto0,c(t)\begin{pmatrix}0&-1\\1&0\end{pmatrix} $$ is a smooth geometric rough path for all $c$ smooth with values in $\mathbb R$, and its driving path $t\mapsto0$ takes values in every tubular neighborhood of $f$. The support of the solution to the Stratonovich equation then contains the solution to the corresponding controlled equation, which is $t\mapsto(0,0,c(t))$. Since the map $\mathbf w\mapsto X^w\mapsto (X^{w,1},X^{w,2})=(w^1,w^2)$ is continuous, we know that the intersection of the supports of the conditioned $X$ contains only curves with vanishing first coordinates.

It is not the case that the support is a singleton in general, and in fact I believe the support will be generically full when the dimension is high enough. I think the full picture is a bit subtle, but let me give you some elements to explore what this support could be. As a tl;dr, let me just say that the support of $X$ when $W$ is conditioned to staying in a neighborhood of $f$ contains all the solutions to $(\star)$ below, when $w$ is constrained to the same neighborhood but $\beta$ is an arbitrary smooth curve of antisymmetric matrices.

Rough path theory. Let me first paint the crudest picture of rough path theory. Be aware that I will use slight variations on the standard terminology. Basically, one is interested in the solutions to some controlled equation $$ \frac{\mathrm d}{\mathrm dt}X^w_t = \mu(X_t) + \sigma(X_t)\frac{\mathrm d}{\mathrm dt}w_t $$ in $\mathbb R^d$, where $w$ is a given smooth path with values in $\mathbb R^k$. $\mu$ is a vector field on $\mathbb R^d$, and $\sigma$ is a map with values in $L(\mathbb R^k,\mathbb R^d)$, that we assume to be smooth for simplicity.

Namely, a rough path lift of a smooth path is a map of the form $$ (s,t)\mapsto w_t-w_s,\Big(\int_s^t(w^i_u-w^i_s)\frac{\mathrm d}{\mathrm du}w^j_u\mathrm du\Big)_{i,j} $$ with values in $\mathbb R^d\times\mathbb R^{d\times d}$. This is called the (2-step) rough path lift of $w$. We usually write $\mathbb R^d\otimes\mathbb R^d$ for $\mathbb R^{d\times d}$ and $u\otimes v$ for $(u^iv^j)_{i,j}$, for reasons that are extremely natural once one is familiar with tensor algebras. If we write $\mathbf w$ for this pair, the result is that $\mathbf w\mapsto X^w$, which a priori does not depend on the second component, is actually continuous with respect to topologies on the space of lifts of smooth paths that are of practical interest. In some sense, they build up on the $\mathcal C^\alpha$ topology, for $1/3<\alpha<1/2$, but I don't want to go into details; it is however important that for a fixed $\alpha$, it is induced by an explicit metric, and it is stronger than the uniform topology. In particular, it admits a completion $\mathsf{RP}^\alpha([0,1],\mathbb R^d)$ that we can identify with a subset of $\mathcal C^0([0,1]^2,\mathbb R^d\times\mathbb R^d\otimes\mathbb R^d)$. This completion is the set of geometric rough paths of regularity $\alpha$, and the continuous extension of the solution map is called the rough solution map.

Smooth geometric rough paths and support. How does this connect to the problem at hand? Well because of the continuity of the solution map, we know that the support of $X$ when $W$ is conditioned to a tube is the closure of the image of the support of $\mathbf W$ under the same conditioning. However, this is known to contain the lift of any smooth curve with values in the tube, and in fact to be the closure of the set of such lifts. Let us describe some of these elements. If we call smooth geometric rough paths the set of elements $(a,b)$ in $\mathsf{RP}^\alpha$ which are represented by elements of $\mathcal C^\infty([0,1]^2,\mathbb R^d\times\mathbb R^d\otimes\mathbb R^d)$, one can show that they are precisely the smooth maps satisfying the algebraic conditions $$\begin{aligned} a_{s,t} &= a_{s,u}+a_{u,t},\\ b_{s,t} &= b_{s,u}+b_{u,t}+a_{u,t}\otimes a_{s,u},\\ b_{s,t}+b_{s,t}^* &= a_{s,t}\otimes a_{s,t}. \end{aligned}$$ This contains more than just the lifts $\mathbf w$ of smooth paths $w$. Indeed, notice that we can replace $(a,b)$ by $(a,b+\beta)$, where $\beta_{s,t}=\beta_{0,t}-\beta_{0,s}$ and $\beta_{0,t}$ is antisymmetric, and still get a smooth geometric rough path. Hoewever, if $(a,b)$ is a smooth geometric rough path and we define $w:t\mapsto a_{0,t}$, there exists a $\beta$ antisymmetric such that $\beta_{s,t}=\beta_{0,t}-\beta_{0,s}$ and $(a,b)=\mathbf w+(0,\beta)$. In other words, all smooth geometric rough paths arise from adding some sort of internal rotational energy $\beta$ to a smooth path $w$. An important remark is that for every $w$ and $\beta$ as above, there exist smooth curves converging to $w$ in the continuous topology such that their rough path lifts converge to $w+(0,\beta)$. This proves that the support of $X$ conditioned to $W$ staying in some open set of the uniform topology contains the rough solution of the controlled equation driven by all the smooth geometric rough paths $\mathbf w+(0,\beta)$, for $w$ in the open set.

The solution to the controlled equation driven by $\mathbf w+(0,\beta)$ is easy to describe; it is the solution to the ordinary differential equation $$\begin{align*} &(\star)& \frac{\mathrm d}{\mathrm dt}X_t &= \sigma(X_t)\frac{\mathrm d}{\mathrm dt}w_t + \mu(X_t) + \frac12[\sigma,\sigma](X_t)\frac{\mathrm d}{\mathrm dt}\beta_{0,t}, \end{align*}$$ where $[\sigma,\sigma](x)$ is the linear endomorphism sending $u\otimes v$ to the Lie bracket at $x$ of the vector fields $y\mapsto\sigma(y)u$ and $y\mapsto\sigma(y)v$. But the image of $[\sigma,\sigma](x)$ can be much larger than that of $\sigma(x)$.

An example. Let me give an example where we can show that the support is bigger than expected. Choose $k=2$, $d=3$, $\mu\equiv0$, $\sigma(x)u=u^1\partial_{x^1}+u^2\partial_{x^2}+(u^1x^2-u^2x^1)\partial_{x^3}$, $f\equiv0$. Then $$ (s,t)\mapsto0,\big(c(t)-c(s)\big)\begin{pmatrix}0&-1\\1&0\end{pmatrix} $$ is a smooth geometric rough path for all $c$ smooth with values in $\mathbb R$, and its driving path $t\mapsto0$ takes values in every tubular neighborhood of $f$. The support of the solution to the Stratonovich equation then contains the solution to the corresponding controlled equation, which is $t\mapsto(0,0,c(t))$. Since the map $\mathbf w\mapsto X^w\mapsto (X^{w,1},X^{w,2})=(w^1,w^2)$ is continuous, we know that the intersection of the supports of the conditioned $X$ contains only curves with vanishing first coordinates, so the support (say in the space of continuous curves) is precisely the set of curves with vanishing first coordinates.