$\renewcommand{\p}{\partial}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\ip}[2]{\langle #1,#2\rangle}$The answer is yes. 

Indeed, suppose the contrary: for some $v\in X$, some $\zeta\in\p G(v)$, some sequence $(w_n)$ in $X$ converging to $v$, some real $\ep>0$, and all $n$ we have  
\begin{equation}
	\phi(w_n)\le\phi(v)+\ip\xi{w_n-v}-\ep|w_n-v|, \tag{1}\label{1}
\end{equation}
where 
\begin{equation}
	\xi:=\frac{\zeta}{F'(\phi(v))}. 
\end{equation}
The condition $F'>0$ implies that $F$ is increasing. So, \eqref{1} and the mean value theorem imply 
\begin{equation}
	F(\phi(w_n))\le F(\phi(v)+\ip\xi{w_n-v}-2\ep|w_n-v|) \\ 
	=F(\phi(v))+F'(c_n)(\ip\xi{w_n-v}-2\ep|w_n-v|)
\end{equation}
for some sequence $(c_n)$ converging to $\phi(v)$. 
Also, since $F$ is convex and differentiable, we see that $F'$ is continuous and hence $F'(c_n)\to F'(\phi(v))$. So, without loss of generality for all $n$ we have $F(\phi(w_n))\le F(\phi(v))+\ip\zeta{w_n-v}-\ep|w_n-v|$, that is,  
\begin{equation}
	G(w_n)\le G(v)+\ip\zeta{w_n-v}-\ep|w_n-v|,
\end{equation}
which contradicts the condition $\zeta\in\p G(v)$. $\quad\Box$