If by "future null cone" you mean the (achronal) boundary $\partial I^+(p)$ of the chronological future $I^+(p)$ of $p$, then the answer is (almost) yes even without the nontrapping condition, thanks to your smoothness assumption. What may fail is that tangent vectors will be either spacelike or null, and the latter case cannot be excluded without additional assumptions on $\partial M$. For example, take $M$ as the Minkowski space-time minus an open cylinder around the time axis - $S$ will have a null tangent vector at any point where a null generator of $\partial I^+(p)$ meets $\partial M$ tangentially. Of course, this example has trapped null geodesics, but even with the nontrapping assumption the null generators could in principle meet $\partial M$ tangentially, causing $S$ to have null tangent vectors at the meeting point.
The argument goes as follows. Since $\partial I^+(p)$ is assumed to be smooth away from $p$ and this point is assumed to be in the interior of $M$, it turns out that $\partial I^+(p)$ meets $\partial M=\mathbb{R}\times\partial M_0$ transversally. This is due to the fact that timelike boundary curves in $\partial M$ such as $t\mapsto(t,p_0)$, $p_0\in\partial M_0$ are never tangent to $\partial I^+(p)$ (tangent vectors to a lightlike hypersurface are either lightlike or spacelike). As a consequence, $S=\partial I^+(p)\cap\partial M$ is indeed a smooth hypersurface of $\partial M$ (hence with codimension two in $M$) if nonvoid, and $TS$ has no timelike vectors. Moreover, also due to the assumed smoothness of $\partial I^+(p)$ away from $p$, each $q\in S$ belongs to exactly one null generator $\gamma_q$ of $\partial I^+(p)$. If $\gamma_q$ meets $\partial M$ transversally at $q=\gamma_q(t)$, then $\dot{\gamma}_q(t)$ is not tangent to $S$ at $q$ and therefore $T_q S$ is spacelike. If, on the other hand, $\gamma_q$ meets $\partial M$ tangentially at $q$, then $\dot{\gamma}_q(t)$ is tangent to $S$ and therefore $T_q S$ is null.
To rule out $S$ having null tangent vectors, a sufficient condition is that $\partial M$ is totally geodesic. More generally, one could impose a sort of "convexity" assumption on $\partial M$ with respect to null geodesics.