If $Z$ is reduced, then the answer is yes, and the proof is rather elementary.
First, we can suppose $X$ is affine. Indeed, cover $X$ by affine open subsets $X_i$'s. Then $Z\cap X'_i$ is also stable by $F'_X$ because both $X'_i$ and $Z$ are stable by $F'_X$. If we can show that $Z\cap X'_i$ is defined over $k$, then $Z$ is clearly defined over $k$ too.
So suppose $X=\mathrm{Spec} A$ is affine. The closed subscheme $Z$ is defined over some finite extension $K/k$ of degree $r$. Let $I$ be the defining ideal of $Z$ in $X_K$. We have to show that $I$ is stable by $\mathrm{Gal}(K/k)$. Consider the generator $\sigma: \lambda \mapsto \lambda^{q^{r-1}}$ of $\mathrm{Gal}(K/k)$. Let $f=\sum_i b=\sum_i a_i\otimes \alpha_i\in I$ with $a_i\in A$ and $\alpha_i\in K$. Denote by $F_{X_K}$ (F_{X})_K$ the relative Frobenius $F_X$ of $X$ extended to $K$. Then $$\sigma(f)^q=(\sum_{i} $\sigma(b)^q=(\sum_{i} a_i\otimes \alpha_{i}^{q^{r-1}})^q=\sum_{i} a_i^q\otimes \alpha_{i}=(F_{X})_K(f)\in alpha_{i}=(F_{X})_K(b)\in I$$ Therefore $\sigma(f)\in \sigma(b)\in \sqrt{I}=I$ and we are done.
