Yes, this is true, though your particular map $j$ may not come from a $\overline{\mathbb{Q}}$-map (for example, let $X=C\times \mathbb{P}^1$ where $C$ is a $\overline{\mathbb{Q}}$-curve, and let $j$ be the inclusion of a fiber over a $\mathbb{C}$-point of $C$ which is not an $\overline{\mathbb{Q}}$-point).

Suppose $X$ is defined by some homogeneous polynomials $f_1(\underline{x}), \cdots, f_n(\underline{x})$ in variables $\underline{x}=(x_0, \cdots, x_r)$. The map $\mathbb{P}^1$ to $X$ is given by rational functions $\underline{s}=(s_0(t), \cdots, s_r(t))\in \mathbb{C}(t)$ so that $f_i(\underline{s})=0$. Now the $$s_i(t)=\frac{\sum a_{ij}t^j}{\sum b_{il} t^l},$$
where the $a_{ij}\in \mathbb{C}$ (satisfying some conditions, e.g. the $b_{il}$ are not all zero).

Consider the subvariety of $\mathbb{A}^N=\operatorname{Spec}(\overline{\mathbb{Q}}[a_{ij}, b_{il}]_{i,j,l}$) defined by the equations $\{f_r(\underline{s})=0\}$, viewed as equations in the $a_{ij}, b_{il}$. The assumption of the existence of $j$ shows that it is non-empty. The conditions of defining a map (e.g. the $b_{il}$ are not all zero, etc.) which is a closed embedding are open conditions. (Showing that the space of parameters defining a closed embedding is open is non-trivial, so I'll expand on it on request). Thus we have a non-empty open set parametrizing maps of the desired form. So it contains a $\overline{\mathbb{Q}}$-point, completing the proof.

Let me remark that it is much easier to prove this if you black-box the existence of the Hilbert scheme ${\operatorname{Hilb}}(X)$, which is locally of finite type over $\overline{\mathbb{Q}}$. Then one uses the fact that $\mathbb{P}^1$ does not deform to argue that in an open neighborhood of the point of $\operatorname{Hilb}(X)$ representing $j$, we have subschemes isomorphic to $\mathbb{P}^1$. But perhaps it is illuminating to do things explicitly.