The answer is no.
Indeed, suppose that $n=1$ and $f(x)=e^{-x}$. Then $$X(t)=\ln\left(-\frac{1}{4} \left(-\sinh \left(\sqrt{2} t\right)-\cosh \left(\sqrt{2} t\right)-1\right)^2 \left(\sinh \left(\sqrt{2} t\right)-\cosh \left(\sqrt{2} t\right)\right)\right),$$ so that $X(t)\to\infty$ as $t\to\infty$, and also the velocity $X'(t)=\sqrt{2} \tanh \left(\dfrac{t}{\sqrt{2}}\right)$ is increasing with $t$.
Here is the graph $\{(t,X(t))\colon0\le t\le10\}$:
