$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\epi}{\operatorname{epi}}$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. $\quad\Box$
$\newcommand\ep\varepsilon$First, the conditions that $f_n\in\mathcal{C}^1([0,1],\mathbb{R})$ and $f_n(x)\ge\sqrt{x}$ for $x\in[0,1]$ imply $f_n(0)>0$. Since
\begin{equation*}
\begin{cases}
y_n(0)=0, \\
y_n'=f_n(y_n) \text{ on [0,1]},
\end{cases}
\tag{2}\label{2}
\end{equation*}
we see that $y_n>0$ in a right neighborhood of $0$. Since $y'_n=f_n(y_n)\ge0$, we see that $y_n>0$ on $(0,1]$.
Letting then $u_n:=\sqrt{y_n}$, we get $2u_n u'_n=f_n(u_n^2)\ge u_n$, whence $u'_n\ge1/2$, $u_n(x)\ge x/2$, and
\begin{equation*}
y_n(x)\ge x^2/4 \tag{3}\label{3}
\end{equation*}
for all $x\in[0,1]$.
Next, for $\ep\in(0,1)$, let $z_\ep$ be the unique solution of the ODE
\begin{equation*}
z'_\ep=\sqrt{z_\ep}+\ep
\end{equation*}
on $[0,1]$ with the initial condition $z_\ep(0)=0$. It is not hard to see that
\begin{equation*}
z_\ep(x)\to x^2/4 \tag{4}\label{4}
\end{equation*}
uniformly in $x\in[0,1]$ as $\ep\to0$. (See details on this at th end of this answer.)
Let now
\begin{equation*}
\ep_n:=\sup_{x\in[0,1]}\big|f_n(x)-\sqrt{x}\big|
=\sup_{x\in[0,1]}\big(f_n(x)-\sqrt{x}\big),
\end{equation*}
so that $\ep_n\to0$ (as $n\to\infty$), and then let
\begin{equation*}
w_n:=z_{\ep_n+1/n}.
\end{equation*}
So, $y_n(0)=0=w_n(0)$,
\begin{equation*}
y_n'\le\sqrt{y_n}+\ep_n,\quad w_n'=\sqrt{w_n}+\ep_n+1/n.
\end{equation*}
Suppose that
\begin{equation*}
x_n:=\sup\{x\in[0,1]\colon y_n\le w_n\text{ on }[0,x]\}<1.
\end{equation*}
Then $x_n>0$ and $w_n(x_n)=y_n(x_n)$, and hence $y'_n(x_n)\ge w'_n(x_n)$, so that
\begin{equation*}
\sqrt{w_n(x_n)}+\ep_n=\sqrt{y_n(x_n)}+\ep_n \\
\ge y'_n(x_n)\ge w'_n(x_n)=\sqrt{w_n(x_n)}+\ep_n+1/n,
\end{equation*}
a contradiction. So, $x_n=1$ and hence, in view of \eqref{3},
\begin{equation*}
x^2/4\le y_n(x)\le w_n(x)=z_{\ep_n+1/n}(x)\to x^2/4
\end{equation*}
uniformly in $x\in[0,1]$, by \eqref{4}.
On the other hand, the only solution $y$ of the system
\begin{equation*}
\begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}
\end{equation*}
such that $y>0$ on $(0,1]$ is given by the formula $y(x)=x^2/4$.
Thus, $y_n\to y$ uniformly on $[0,1]$, as desired.
Details on \eqref{4}: Letting $t_\ep:=\sqrt{z_\ep}$, rewrite
$z'_\ep=\sqrt{z_\ep}+\ep$ as $2t_\ep t'_\ep=t_\ep+\ep$. "Separating the variables", we find that $t_\ep=g^{-1}_\ep$, where
\begin{equation*}
g_\ep(t):=2t-2\ep\ln\frac{t+\ep}\ep.
\end{equation*}
We have $g'_\ep(t):=2-\frac{2\ep}{t+\ep}>0$ for $t>0$, so that the inverse function $g^{-1}_\ep$ is well defined. Using the inequalities $\ln(t+\ep)\le t+\ep-1<t$, we get
\begin{equation*}
g_\ep(t)\ge (2-2\ep)t+2\ep\ln\ep
\end{equation*}
for $t\ge0$, whence for $\ep\to0$ we have $t_\ep(x)=g^{-1}_\ep(x)\le\frac{x-2\ep\ln\ep}{2-2\ep}\to x/2$ and
\begin{equation*}
z_\ep(x)=t_\ep(x)^2\le(1+o(1))x^2/4
\end{equation*}
uniformly in $x\in[0,1]$. Also, similarly to \eqref{3}, $z_\ep(x)\ge x^2/4$ for $x\in[0,1]$. Now \eqref{4} follows.
Best Answer
For any $x_0\in K$ we have the trajectry $x(t)$ which go out of $K$ ,say, at time $t_0=\tau^K(x_0)+\varepsilon$. Hence, by the continuous dependence on its initial value, the solution starts from an neighborbood $U(x_0)$ of $x_0$ also goes out of $K$ at $t=t_0$. Covering the compact set $K$ by finite number of such $U(x_0)$'s you have the uniform bound of the exit time.