I am posting my comments as an answer.  I am concerned that I misunderstand the OP, so let me state first the result.  There exists a finite field extension $K/\mathbb{Q}$ such that for every closed immersion $\mathbb{P}^1_\mathbb{C} \hookrightarrow X\otimes_{\mathbb{Q}}\mathbb{C}$, there exists a closed immersion $\mathbb{P}^1_K \hookrightarrow X \otimes_{\mathbb{Q}} K$ with the same image after base change to $X\otimes_{\mathbb{Q}}\mathbb{C}$.  

To prove this, first observe that the geometric Picard group $\text{Pic}(X\otimes_{\mathbb{Q}}\mathbb{C})$ is finitely generated.  Thus, there exists a finite field extension after which a finite generating set of the geometric Picard group is defined over that field.  

Second, the geometric automorphism group of the complex scheme, $\text{Aut}_{\mathbb{C}}(X\otimes_{\mathbb{Q}}\mathbb{C})$, is a discrete group that is finitely generated by Corollary 2.4 of Chapter 15, p. 315 of the following; result due to H. Sterk -- Finiteness results for algebraic K3 surfaces. Math. Z., 189(4):507–513, 1985.   

Daniel Huybrechts <br>
<I>Lectures on K3 Surfaces</I> <br>
Part of Cambridge Studies in Advanced Mathematics <br>
September 2016 <br>
ISBN: 9781107153042 <br>
http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf

Thus, after a further finite field extension, there exists a finite subset of $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ whose images generate the geometric automorphism group.  Thus, $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ is already the full geometric automorphism group.

Third, by Corollary 4.6 of Chapter 8, p. 161 -- result again proved by Sterk in the article cited above -- the action of the geometric automorphism group on the set of smooth rational curves has only finitely many orbits.  For each of those finitely many orbits, after a degree $2$ field extension, every curve in that orbit acquires a rational point.  Thus, after a further finite field extension, every smooth, genus $0$ curve on $X$ acquires a rational point.

Please note, all of this is valid for every K3 surface, not only for the Fermat quartic surface.