$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\epi}{\operatorname{epi}}$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. $\quad\Box$
The equality
$$T_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$
is false in general, because $T_K(u)$ is closed by definition, whereas
$$S_K(u):=K-\mathbb R_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$
can be not closed.
A counterexample is provided by this answer. Indeed, let
$$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$
and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.
Also, $w:=(0,1,0)\notin S_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting
$$w_t:=v_t-tu$$
for real $t>0$ and $v_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v_t\in K$ and hence $w_t\in K-\mathbb R_+ u=S_K(u)$, whereas $w_t\to w$ as $t\to\infty$. $\quad\Box$
Best Answer
$\newcommand{\R}{\mathbb R}\newcommand{\inter}{\operatorname{int}}$The answer is yes.
Here is the proof.
The asymptotic cone of $C$ is \begin{equation*} K:=C^{as}=\{y\in\R^n\colon \exists(x_k) \text{ in }C\ \;\exists(t_k) \text{ in }\R\ \; t_k\to\infty\ \&\ x_k/t_k\to y\}. \tag{10}\label{10} \end{equation*} The polar cone to $K$ is \begin{equation*} K^\circ:=\{z\in\R^n\colon z\cdot y\le0\ \forall y\in K\}, \end{equation*} where $\cdot$ denotes the dot product.
Take any \begin{equation*} z\in\inter(K^\circ). \end{equation*} We want to show that \begin{equation*} z\overset{\text{(?)}}\in Z:=\bigcup_{x\in\partial C} \R_+\nabla f(x), \tag{20}\label{20} \end{equation*} where $\R_+w:=\{tw\colon t\in[0,\infty)\}$ for $w\in\R^n$. Clearly, without loss of generality (wlog) \begin{equation} z\ne0. \tag{30}\label{30} \end{equation}
Proof: Take any $y\in K\setminus\{0\}$. Then $z+ty\in K^\circ$ for all small enough real $t>0$, so that $z\cdot y=(z+ty)\cdot y-ty\cdot y<(z+ty)\cdot y\le0$. $\quad\Box$
Proof: Suppose the contrary, that the supremum of the set $z\cdot C$ is not attained. Let $(x_k)$ be a sequence in $C$ such that $z\cdot x_k\to s$.
If the sequence $(x_k)$ is bounded, then, passing to a subsequence, we see that wlog $x_k\to x^*$ for some $x^*\in\R^n$; moreover, $x^*\in C$, since the set $C$ is closed. So, $z\cdot x^*=\lim_k z\cdot x_k=s$, so that the supremum of the set $z\cdot C$ is attained, a contradiction.
So, the sequence $(x_k)$ is unbounded. So, passing to subsequences, we see that wlog $|x_k|\to \infty$ and then $x_k/|x_k|\to y$ for some $y\in\R^n\setminus\{0\}$, where $|\cdot|$ denotes the Euclidean norm. So, by \eqref{10}, $y\in K\setminus\{0\}$, and
\begin{equation} z\cdot y=\lim_k\frac{z\cdot x_k}{|x_k|}\ge0, \end{equation} since $z\cdot x_k\to s\in(-\infty,\infty]$ and $|x_k|\to\infty$. So, we get a contradiction with Lemma 1. So, the supremum of the set $z\cdot C$ is attained at the point $x^*\in C$. Finally, in view of \eqref{30}, the supremum of the set $z\cdot C$ cannot be attained at any point in $\inter C$. So, Lemma 2 is proved. $\quad\Box$
Let now $h:=\nabla f(x^*)$, so that $h\ne0$, by a condition in the OP. If $z\notin\R_+ h$, then there is some $y\in\R^n$ such that $y\cdot h<0<y\cdot z$. So, the directional derivative $y\cdot h$ of $f$ at $x^*$ in the direction of $y$ is $<0$. So, for all small enough real $t>0$ we have $f(x^*+ty)<f(x_*)\le0$ and hence $x^*+ty\in C=\{x\in\R^n\colon f(x)\le0\}$, which implies \begin{equation} z\cdot x^*=s\ge z\cdot(x^*+ty)=z\cdot x^*-tz\cdot y<z\cdot x^*, \end{equation} a contradiction.
Thus, \begin{equation} z\in\R_+ h=\R_+\nabla f(x^*)\subseteq Z, \end{equation} so that \eqref{20} holds. $\quad\Box$
One may note that, instead of the conditions $f\in\mathcal{C}^2(\R^n;\R)$ and $\forall x\in\R^n$ $\nabla f(x)\ne0$ in the OP, in the above proof we only used the weaker restriction that $f$ is differentiable on $\partial C$ with $\nabla f(x)\ne0$ for $x\in\partial C$.
Moreover, the condition that the Hessian of $f$ be positive definite everywhere was not used at all. Furthermore, note that the positive definiteness of the Hessian of $f$ implies that the set $C$ is convex -- but even the convexity of $C$ was not used in the proof.
The latter remark is illustrated in the picture below, showing (for $n=2$) the (nonconvex, light-gray) set $C\cap\big([0,2\sqrt c\,]\times[-2\sqrt c,2\sqrt c\,]\big)$ for $f=f_c$ and $c=8$, where \begin{equation} f_c(X):=g_c(y)-x,\quad g_c(y):=\sqrt{y^2+\frac{c}{1+c y^2}} \end{equation} for $X=(x,y)\in\R^2$. Shown also are the corresponding (yellow) set $K^\circ\cap\big([0,2\sqrt c\,]\times[-2\sqrt c,2\sqrt c\,]\big)$ and (orange) set $Z\cap\big([0,2\sqrt c\,]\times[-2\sqrt c,2\sqrt c\,]\big)$.
In this case, $K=\{(x,y)\in\R^2\colon|y|\le x\}$, $K^\circ=\{(x,y)\in\R^2\colon|y|\le|x|,\,x\le0\}$, $\inter(K^\circ)=\{(x,y)\in\R^2\colon|y|<|x|,\,x<0\}$, and $$Z=\{(x,y)\in\R^2\colon|y|\le k_8|x|,\,x\le0\},$$ where $k_8:=\max_{y\in\R}g'_8(y)=2.870\dots$, so that $\inter(K^\circ)\subsetneq\inter Z$.