$\newcommand\la\lambda$No. E.g., if $n=1$, $x_0\ne0$, and $f(u)=-u^2/2$ for all real $u$, then the solution
$$x(t)=\frac{x_0}{2} \, \Big(\frac{e^{\la_+ t}-e^{\la_- t}}{\sqrt{4 g+1}}+e^{\la_- t}+e^{\la_+ t}\Big)$$
of the problem
$$\ddot{x}+\dot{x}+g\nabla f(x)=0, \quad x(0)=x_0,\quad\dot{x}(0)=0 \tag{1}\label{1}$$
will be bounded in $t\ge0$ if $0<g\le2$ and unbounded in $t\ge0$ if $g>2$, where
$$\la_\pm:=\frac{-1\pm\sqrt{4 g+1}}2.$$

**Addendum:** The OP later added the condition that $f$ be bounded below. This changes the problem dramatically.

First here, note that, if $x(t)$ is the solution to \eqref{1}, then for the "energy"
$$E(t):=\frac{|\dot x(t)|^2}2+gf(x(t))$$
and all real $t\ge0$ we have
$$E'(t)
=(\ddot x(t)+g\,\nabla f(x(t))\cdot\dot x(t)
=-\dot x(t)\cdot\dot x(t)
=-|\dot x(t)|^2\le0,$$
so that the "energy" is nonincreasing; of course, here $|\cdot|$ is the Euclidean norm and $\cdot$ is the dot product. It follows that for all real $t\ge0$
$$f(x(t))\le c:=E(0)/g<\infty.$$

If now $f$ is coercive (that is, $f(x)\to\infty$ as $|x|\to\infty$), then the set $\{y\in\mathbb R^n\colon f(y)\le c\}$ is bounded, and hence the solution $x(t)$ is bounded (in real $t\ge0$).

In a comment, the OP said "we can assume $f$ to be polynomial".

If $n=1$, then any bounded from below non-constant polynomial is coercive, and the case of a constant polynomial is quite easy. So, this solves the case $n=1$ (still with a polynomial $f$).

Even if $n>1$, it seems possible to show that "almost all" bounded from below non-constant polynomials are coercive, and for such polynomials, the solution $x(t)$ will be of course bounded.