$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\epi}{\operatorname{epi}}$Yes, $p$ is Lipschitz.
Indeed,
\begin{equation*}
p(x)=x+f(x)u
\end{equation*}
for $x\in C$, where
\begin{equation*}
f(x):=\inf E_x,\quad E_x:=\{t\ge0\colon x+tu\notin C\}.
\end{equation*}
So, it suffices to show that the function $f$ is Lipschitz on $C$.
Let us use the Rockafellar, p. 62 notation $0^+C$ for $C^{as}$ and then $(0^+C)^\circ$ for the interior of $0^+C$.
For each $x\in C$, we have
\begin{equation*}
f(x)=\infty\iff E_x=\emptyset\iff (\forall t\ge0\ x+tu\in C)\iff u\in0^+C;
\end{equation*}
the latter $\iff$ follows by Rockafellar, Theorem 8.3, p. 63. So, in view of the condition $u\notin0^+C$, we see that
\begin{equation}
\text{$0\le f<\infty$ on $C$. } \tag{0}\label{0}
\end{equation}
Without loss of generality (wlog), the vector $u$ is of unit length.
Since
\begin{equation*}
v:=-u\in(0^+C)^\circ,
\end{equation*}
there is some real $\ep>0$ such that
\begin{equation*}
v+\ep B\subseteq 0^+C, \tag{1}\label{1}
\end{equation*}
where $B$ is the closed unit ball centered at the origin.
It is enough to show that the restriction of $f$ to the intersection of $C$ with any 2-D affine plane parallel to the vector $u$ is $L$-Lipschtiz, with the same Lipschitz constant
\begin{equation*}
L:=1+l,\quad l:=1/\ep \tag{2}\label{2}
\end{equation*}
for all such 2-D affine planes.
Therefore, wlog the dimension $n$ is $2$. Also then, wlog $-u=v=(0,1)$. It then follows that
\begin{equation*}
f(a,b):=f((a,b))=b-g(a) \tag{3}\label{3}
\end{equation*}
for $(a,b)\in C\subseteq\R^2$, where
\begin{equation*}
g(a):=\inf\{y\in\R\colon(a,y)\in C\}
\end{equation*}
for all real $a$, so that $C$ is the epigraph $\epi(g):=\{(a,y)\in\R^2\colon y\ge g(a)\}$ of the (necessarily convex) function $g$. Here, the standard convention $\inf\emptyset=\infty$ is used (even though we will see in a moment that in our case in fact we have $g(a)<\infty$ for all real $a$).
By \eqref{1} and \eqref{2}, $\epi(g)$ contains $\epi(h)$ for a function $h$ given by a formula of the form $h(a)=B+l|a-A|$ for some real $A,B$ and all real $a$. Then $g\le h$; in particular, $g<\infty$ on $\R$. Also, by \eqref{3} and \eqref{0}, $g(a)>-\infty$ for some real $a$. So, for the convex function $g$ we have $g>-\infty$ on $\R$. So, $-\infty<g<\infty$ on $\R$.
Now it follows from $g\le h$ that the limit $g'(\infty)$ of the (say) right derivative $g'(a)$ of the (finite convex) function $g$ as $a\to\infty$ is $\le h'(\infty)=l$. Similarly, the limit $g'(-\infty)$ of the right derivative $g'(a)$ of the convex function $g$ as $a\to-\infty$ is $\ge h'(-\infty)=-l$. Since the right derivative $g$ of the convex function $g$ is nondecreasing, we see that $-l\le g'\le l$. So, $g$ is $l$-Lipschitz.
Thus, by \eqref{3} and \eqref{2}, $f$ is indeed $L$-Lipschtiz. $\quad\Box$
Best Answer
If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).
To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $W\times(t_0-\delta,t_0+\delta)$ of $(x_0,t_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in W\times(t_0-\delta,t_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t_0+\delta))>0$ and $x^1(x(t_0-\delta))<0$, shrinking the neighborhood $W$ if necessary we get that $x^1(y(t_0+\delta))>0$ and $x^1(y(t_0-\delta))<0$ for all $y\in W$. So by the intermediate value theorem, for every $y\in W$ there is some $s\in(t_0-\delta,t_0+\delta)$ such that $y(s)\in W$.
As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)<t_0+\delta\;\forall y\in W$. (We have still not used that $V$ is closed)
Now we just need to prove that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)>t_0-\delta\;\forall y\in W$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<t_0-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq t_0-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.