Edit: I just noticed that the OP asked about a 1-d Brownian motion. The constriction below only works in three or more dimensions. Back to the drawing board....
Your function $V$ is not necessarily continuous. Its continuity properties depend not only on the function $f$, but also the nature of the random time $\tau$.
A classic counterexample is found by letting $\tau$ to be the hitting time of the complement of a bounded, open region $D$ with an irregular point (as defined in Newtonian potential theory). For instance, you could choose a region with a "Lebesgue spine".
http://en.wikipedia.org/wiki/Lebesgue_spine
Then $E_x[\tau]<\infty$ for all $x$, and for any continuous $f$ your function $V(x):=E_x[f(B_\tau)]$ is the Perron-Wiener-Brelot solution to the Dirichlet problem with data $(D,f)$. That is, $V$ is harmonic on $D$ and $\lim_{D\ni x\to z}V(x)=f(z)$ at all regular points $z\in \partial D$.
However, if the point $z$ is irregular, then choosing $f$ with $f(z)=1$ and $f(y)<1$ otherwise, we have $\liminf_{D\ni x\to z}V(x)$<1. On the other hand, $V(y)=f(y)$ for all $y\not\in \bar D$ so approaching the tip of the spine from outside of $\bar D$, the function $h$ V$ has limit 1.
Thus, $V$ fails to be continuous at $z$. Note that $f$ can be as smooth as you like.
Intuitively, the reason why $V$ is discontinuous is that the spine is so sharp that Brownian motion fails to see it, even as the starting point approaches the tip of the spine from within $D$.
One nice treatment of these questions of probabilistic potential theory is found in Kai Lai Chung's "Lectures from Markov Processes to Brownian Motion". The lim inf result above is Theorem 3 (p.164) in section 4.4 of this book.
