The answer is no.
Indeed, suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$
So, $X(t)$ does not converge as $t\to\infty$.
On the other hand, still for $n=1$, if the velocity $X'$ is nondecreasing, then $X''\ge0$ and hence $X$ is convex. So, in view of the condition $X'(0)=0$, $X$ is nondecreasing. It also follows that $f'(X(t))=-X''(t)\le0$. Since $f$ is strongly convex, we have $f'(x)\to\infty$ as $x\to\infty$. So, $X$ is bounded from above. Since $X$ is convex and nondecreasing, it follows that $X$ is constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.