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 $V$ 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.