Here is a counterexample to the convergence of $\nabla F$ in the conditions you give.
Let $f(x)=\sqrt{1+x^2}$, let $s(x)=\cos(\ln(\ln(x^2+10)))$ and let $\phi:\mathbb{R}\to[0,1]$ be some $C^\infty$ bump function with supp$(\phi)\subseteq[-1,1]$ and $\phi(0)=1,\phi'(0)=0$. Then, for a small enough constant $\varepsilon>0$, the function $g:\mathbb{R}^2\to\mathbb{R}$;
$$g(x,y)=f(x)+f(y+\varepsilon\phi(y)s(x))$$
has $\frac{\partial}{\partial y}g(x,0)=f'(\varepsilon s(x))$ (which does not converge when $x\to\infty$). But $g$ is convex, as we will see below by just computing its hessian. We just need to check that the Hessian is positive definite for $|y|\leq1$, as otherwise $g(x,y)=f(x)+f(y)$, which is strictly convex. As $|y|\leq1$, we have $y+\varepsilon\phi(y)s(x)\in[-2,2]$ if $\varepsilon\leq1$.
$$\frac{\partial}{\partial x}g(x,y)=f'(x)+f'(y+\varepsilon\phi(y)s(x))\varepsilon\phi(y)s'(x).
$$
$$\frac{\partial}{\partial y}g(x,y)=f'(y+\varepsilon\phi(y)s(x))(1+\varepsilon s(x)\phi'(y)).$$
Note that $\frac{\partial}{\partial x}g(x,y)$ and $\frac{\partial}{\partial y}g(x,y)$ are bounded.
$$\frac{\partial^2}{\partial x^2}g(x,y)=f''(x)+f''(y+\varepsilon\phi(y)s(x))(\varepsilon\phi(y)s'(x))^2+f'(y+\varepsilon\phi(y)s(x))\varepsilon\phi(y)s''(x)$$
$$\frac{\partial^2}{\partial x\partial y}g(x,y)=f''(y+\varepsilon\phi(y)s(x))\varepsilon\phi(y)s'(x)(1+\varepsilon s(x)\phi'(y))+f'(y+\varepsilon\phi(y)s(x))\varepsilon s'(x)\phi'(y)$$
$$\frac{\partial^2}{\partial y^2}g(x,y)=f''(y+\varepsilon\phi(y)s(x))(1+\varepsilon s(x)\phi'(y))^2+f'(y+\varepsilon\phi(y)s(x))\varepsilon s(x)\phi''(y)$$
Now, for some constant $C$ we have $f''(x)=\frac{1}{(1+x^2)^{3/2}}>C|s''(x)|$ for all $x\in\mathbb{R}$. So by taking $\varepsilon$ to be small enough, we have
$$\frac{\partial^2}{\partial x^2}g(x,y)>\frac{1}{2}f''(x)=\frac{1}{2(1+x^2)^{3/2}}.$$
Also, as $|y|\leq1$ implies $f''(y+\varepsilon\phi(y)s(x))>0.01$, taking $\varepsilon$ small enough we have
$\frac{\partial^2}{\partial y^2}g(x,y)>0.001$.
Meanwhile, for some constant $K$, the term $\frac{\partial^2}{\partial x\partial y}g(x,y)$ is bounded in norm by $K\varepsilon s'(x)\leq\frac{2K\varepsilon}{\sqrt{x^2+10}}$. From these estimations it follows that the trace and determinant of the Hessian of $g$ are positive, so the Hessian of $g$ is positive definite, so $g$ is convex.
