Here is a sketch of an argument under some additional assumptions (based on discussions with Jacques Audibert).
First of all, observe that $\Gamma \subseteq G(\mathbb{R})$ is discrete. Indeed, it is sufficient to check that $\{e\}$ is an isolated point of $\Gamma\subset G(\mathbb{R})$ (in the analytic topology). Suppose that there exists a sequence of elements $(\gamma_n)$ in $\Gamma$ that tends to the unity. Since $X(\mathbb{Z}) \subset X(\mathbb{C})$ is discrete in the analytic topology, for $n$ sufficiently big the element $\gamma_n$ acts trivially on $X(\mathbb{Z})$. Since $X(\mathbb{Z})$ is Zariski dense in $X(\mathbb{C})$, this implies that $\gamma_n$ acts trivially on $X(\mathbb{C})$ as well, and $\gamma_n=e$.
Now assume further that:
- $G$ is unimodular;
- the action of $G$ on $X$ admits a linearisation over $\mathbb{Z}$. This means that there exists a faithful representation $\rho \colon G \to \operatorname{GL}_r(\mathbb{Q})$ and an equivariant embedding $j
\colon X(\mathbb{Q}) \hookrightarrow \mathbb{Q}^r$, such that $j(X(\mathbb{Z})) = j(X(\mathbb{Q})) \cap \mathbb{Z}^r$.
Denote $G(\mathbb{Z}):=\rho^{-1}(\rho(G) \cap \operatorname{SL}_r(\mathbb{Z}))$. This is an arithmetic lattice in $G$. Observe that the second condition guarantees that $G(\mathbb{Z}) \subseteq \Gamma$. Since $\Gamma$ is discrete, $\Gamma \backslash G(\mathbb{R})$ is an orbifold, and there is a cover $f \colon G(\mathbb{Z}) \backslash G(\mathbb{R}) \to \Gamma \backslash G(\mathbb{R})$. We have $\operatorname{vol}(G(\mathbb{Z}) \backslash G(\mathbb{R}))=\operatorname{deg}(f) \operatorname{vol}(\Gamma \backslash G(\mathbb{R}))$, which implies that $\Gamma$ is a lattice and $G(\mathbb{Z})$ is of finite index in $\Gamma$. Thus, $\Gamma$ is arithmetic.
Remark. I am not sure how restrictive the second condition is; there is a simple example, showing that it is at least not completely empty: Let $G=\mathbb{G}_m$ and $X=\mathbb{A}^1 \setminus \{0\}$. Then the standard linearisation of the action via $\rho \colon \mathbb{G}_m \to \operatorname{GL}_2(\mathbb{Q}), \ t \mapsto \begin{pmatrix} t & 0 \\ 0& t^{-1} \end{pmatrix}$ and $j \colon X \to \{uv=1\} \subset \mathbb{A}^2, \ x \mapsto (x, x^{-1})$ is not over $\mathbb{Z}$ in this sense.
On the other hand, the construction above still produces $G(\mathbb{Z})=\Gamma$ in this example.
