That is often impossible. I am pretty certain all of the following has been explained here before, but here it is again.

Begin with a smooth, projective $k$-curve $P$ such that $\overline{P}$ is isomorphic to $\mathbb{P}^1_{\overline{k}}=\text{Proj}\ \overline{k}[s,t]$, yet $P$ is not isomorphic to $\mathbb{P}^1_k$, e.g., for $k=\mathbb{R}$, let $P$ be the smooth plane conic in $$\mathbb{P}^2_{\mathbb{R}} = \text{Proj} \ \mathbb{R}[x,y,z]$$ with defining equation $x^2 + y^2 + z^2$. Next, let $s$ be any nonzero global section in $H^0(P,\omega_{P/k}^\vee)$. In the example, by adjunction, $\omega_{P/k}^\vee$ equals $\mathcal{O}_{\mathbb{P}^2_{\mathbb{R}}}(1)|_P$, so let $s$ be the global section of $\mathcal{O}_{\mathbb{P}^2_{\mathbb{R}}}(1)$ corresponding to $z\in \mathbb{R}[x,y,z]_1$. Let $X$ be $D(s)$, the open subscheme of $P$ on which $s$ is a generator of $\omega_{P/k}^\vee$, i.e., in the example $X$ is isomorphic to $\text{Spec} \mathbb{R}[u,v]/\langle u^2 + v^2 +1 \rangle$, where $u=x/z$ and $v=y/z$.

There exists an isomorphism
$$i: \mathbb{P}^1_{\overline{k}} \to \overline{P}$$ with $i^{-1}(\overline{D})$ equal to $D(st)$. Consider the finite flat morphism, $$F:\mathbb{P}^1_{\overline{k}} \to \mathbb{P}^1_{\overline{k}}, \ F^*(s)=s^2, \ F^*(t) = t^2.$$ Denote $F^{-1}(D(st))$ by $Y$.
Assuming that $\text{char}(k)\neq 2$, the induced morphism,
$$ F|_Y : Y \to D(st),$$ is finite and étale. Thus the composition $i\circ F|_Y$ is finite and étale. Denote this by $f$.

By way of contradiction, assume that there exists a $k$-morphis,
$$f^0:Y^0\to X$$
whose basechange to $\overline{k}$ is isomorphic, as a scheme over $\overline{X}$ to $f=i\circ F|_Y$.
By the theory of smooth curves over a field, there is a dense open immersion of $Y^0$ in a smooth, projective $k$-curve $Q^0$, and this open immersion is unique up to unique $k$-isomorphism. Moreover, by the valuative criterion of properness, the morphism $f^0$ extends to a $k$-morphism,
$$ F^0:Q^0 \to P,$$ whose restriction to $Y_0$ equals $f^0$. Since $Y^0$ is dense in $Q^0$, and since $Q^0$ is separated, also the base change of $F^0$ to $\overline{k}$ is equivalent, as a scheme over $\overline{P}$, to $i\circ F$.

In particular, consider the coherent $\mathcal{O}_P$-module $\mathcal{L}$ that is the cokernel of
$$ F_0^\#:\mathcal{O}_P \to F^0_*\mathcal{O}_{Q^0}.$$
The base change of $\mathcal{L}$ to $\mathcal{k}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1_{\overline{k}}}(-1)$. Thus, $\mathcal{L}^\vee$ is an invertible sheaf on $P$ whose base change is isomorphic to $\mathcal{O}_{\mathbb{P}^1_k}(1)$. But then the complete linear system of $\mathcal{L}^\vee$ defines a $k$-isomorphism of $P$ to $\mathbb{P}^1_k$, contrary to hypothesis. Therefore, there does not exist any $k$-morphism $f^0$ whose basechange to $\overline{k}$ equals $f$.