Edit: This question has a solution here: https://mathoverflow.net/a/402426
Suppose that $f: S^2\rightarrow \mathbb{C}P^1$ is a continuous map from a topological $2$-sphere to the Riemann sphere which is a local homeomorphism away from isolated points (this may be too weak, see example below for what I am thinking of). Suppose that $f$ gives us a complex structure on $S^2$ by using charts $(\phi\circ f,f^{-1}(U))$ away from branch points of $f$, and some appropriate root of $f$ or $1/f$ at the isolated points where $f$ is ramified. With this new complex structure on $S^2$, there exists a uniformising parameter $z$ (up to Mobius transformations) for which we can write $f(z)$ as a holomorphic rational function. (this is similar to dessin d'enfants (Wikipedia: see section 2 paragraph 4))
Here is a (sort of tautological) example. suppose $S^2=\mathbb{C}P^1$ and $f(z) = (1 + z^2)^3/(3z/2 + z^3)^2$ (this is a Belyi function). Then 'forgetting' the complex structure on $S^2$, we can use $f$ to give $S^2$ a complex structure in which the uniformising parameter $z$ gives $f$ the form $f(z) = z^2/((z-2)(z-1))$.
Question: If I replace $f$ above with a continuous family $f_t$ with $f_0=f$, then will the corresponding holomorphic maps $f_t(z)$ also be a continuous family of holomorphic maps in a single variable $z$ (at least for small $t$)?
The step I am worried about is the derivation of the uniformising parameter since I do not have a good understanding of the uniformisation theorem. How do I know that the roots and poles of the resulting holomorphic functions $f_\alpha$ do not 'jump' in a discontinuous way? I may be overthinking this, and it might follow immediately form the fact that $f_t$ is continuous in $t$.