No. Even if $X$ is "defined over $\overline{\mathbb{Q}}$" in the sense that $X$ is isomorphic to $X_0\otimes_{\overline{\mathbb{Q}}} \mathbb{C}$ for some variety $X_0$ over $\overline{\mathbb{Q}}$, nonetheless the stack $\mathcal{X}$ may not be defined over $\overline{\mathbb{Q}}$.

For instance, let $X_0$ be $\mathbb{P}^1_{\overline{\mathbb{Q}}}$ so that $X$ is $\mathbb{P}^1_{\mathbb{C}}$. Now let $t_0,t_1,t_2\in \mathbb{P}^1_{\mathbb{C}}(\mathbb{C})\setminus\{0,1,\infty\}$ be three distinct points such that at least one is transcendental, e.g., $2$, $3$, $\pi$. Denote by $$f:Y\to X,$$ the genus $2$, hyperelliptic curve branched over $$\{0,1,\infty,t_0,t_1,t_2\}.$$ There is an action of $\mathbb{Z}/2\mathbb{Z}$ on $Y$ via the hyperelliptic involution. Let $\mathcal{X}$ be the quotient Deligne-Mumford stack, $$\mathcal{X} = [Y /(\mathbb{Z}/2\mathbb{Z})].$$ The coarse moduli space is $X$, via the unique morphism $\pi: \mathcal{X}\to X$ that factors $f$. Of course $X$
is defined over $\overline{\mathbb{Q}}$. However, the branch divisor of the morphism $\pi$ is not defined over $\overline{\mathbb{Q}}$ as a subscheme of $X$. Thus, also $\mathcal{X}$ is not defined over $\overline{\mathbb{Q}}$.

Added by OP:

It might be useful to (slightly) generalize the above construction.

Let $X=\mathbb P^1_{\mathbb C}$ and let $f:Y\to X$ be a hyperelliptic curve with branch locus $D$ (and $Y$ of positive genus). Write $U=X\backslash D$. Let $\mathcal X$ be the stack $[Y/G]$. Then the unique morphism $\pi:\mathcal X \to X$ that factors $f:Y\to X$ is the coarse moduli space of $\mathcal X$. Suppose that $U$ can't be defined over Qbar. Then $Y\to X$ (and $Y$) can't be defined over $\overline{\mathbb Q}$ (as the hyperelliptic involution is unique). Therefore, $\mathcal X\to X$ and $\mathcal X$ can't be defined over Qbar.