$\newcommand\p\partial\newcommand\R{\mathbb R}\newcommand\cl{\operatorname{cl}}\newcommand\conv{\operatorname{conv}}$The answer is yes, and you were almost there.

Indeed, suppose the contrary: that we we have a continuous function $f\colon\R^n\to\R^n$ such that $f(z)\in\p F(z)$ for all $z\in\R^n$ and yet
$F$ is not differentiable at some $x\in\R^n$.

By Theorem 25.1 in Rockafellar's book, the non-differentiability of $F$ at $x$ means that the cardinality $|\p F(x)|$ of the subdifferential $\p F(x)$ of $F$ at $x$ is $>1$. The subdifferential $\p F(x)$ is a closed convex set. Moreover, in this case $\p F(x)$ is bounded, since the function $F$ is continuous and hence locally bounded. So, the subdifferential $\p F(x)$ contains two distinct extreme points, say $u$ and $v$.

By Theorem 25.6 in Rockafellar's book, $\p F(x)=\cl\conv S(x)+K(x)$, where $\cl\conv S(x)$ is the closed convex hull of the set $S(x)$ of all limits of the sequences of the form $(\nabla F(x_k))$ such that $F$ is differentiable at all $x_k$'s and $x_k\to x$ (as $k\to\infty$), and $K(x)$ is the normal cone to $\operatorname{dom}F$ at $x$. In this case, $K=\{0\}$, since $\operatorname{dom}F=\R^n$. Also, the set $S(x)$ is closed and bounded, again because the function $F$ is locally bounded. So, $\p F(x)=\cl\conv S(x)$ and hence any extreme point of $\p F(x)$ is in $S(x)$. So, the two distinct points $u$ and $v$ in $\p F(x)$ are in $S(x)$. So, $u=\lim_k \nabla F(y_k)$ and $v=\lim_k \nabla F(z_k)$ for some sequences $(y_k)$ and $(z_k)$ converging to $x$ such that $F$ is differentiable at all $y_k$'s and at all $z_k$'s.

But, again by Theorem 25.1 in Rockafellar's book, $\nabla F(y_k)=f(y_k)$ and $\nabla F(z_k)=f(z_k)$. So, $u=\lim_k\nabla F(y_k)=\lim_k f(y_k)=f(x)$ and $v=\lim_k\nabla F(z_k)=\lim_k f(z_k)=f(x)$, which contradicts the condition that $u$ and $v$ are distinct. $\quad\Box$