Here is a counterexample in which $X/G$ is also Hausdorff.

We give $X:=(\mathbb{N} \times \mathbb{Z}) \cup \{\infty\}$ a topology similar to the Arens-Fort topology. That is, each point $(m,n) \in \mathbb{N} \times \mathbb{Z}$ is isolated and basic neighborhoods of $\infty$ are of the form $$B_{f,k}:=\{\infty\} \cup \{(m,n) : n \geq f(m) \land m \geq k\},$$ where $f:\mathbb{N} \to \mathbb{Z}$ and $k \in \mathbb{N}$.

We let $G:=\mathbb{Z}$ act on $X$ in the natural way: $g(m,n)=(m,n+g)$ and $g(\infty)=\infty$.

Any compact subset of $X$ is finite, so its orbit cannot cover $X$. On the other hand $X/G$ is just a convergent sequence (which is compact and Hausdorff).

Here is a counterexample in which $X/G$ is also Hausdorff.

We give $X:=(\mathbb{N} \times \mathbb{Z}) \cup \{\infty\}$ a topology similar to the Arens-Fort topology. That is, each point $(m,n) \in \mathbb{N} \times \mathbb{Z}$ is isolated and basic neighborhoods of $\infty$ are of the form $$B_{f,k}:=\{\infty\} \cup \{(m,n) : n \geq f(m) \land m \geq k\},$$ where $f:\mathbb{N} \to \mathbb{Z}$ and $k \in \mathbb{N}$.

We let $G:=\mathbb{Z}$ act on $X$ in the natural way: $g(m,n)=(m,n+g)$ and $g(\infty)=\infty$.

Any compact subset of $X$ is finite, so its orbit cannot cover $X$. On the other hand $X/G$ is just a convergent sequence (which is compact and Hausdorff).