The argument I gave in that thread extends to a pretty wide class of surfaces with various symmetries. It's the same argument but there's some facts that are true for the hyperelliptic curves that are a little more fussy for curves with other symmetries. The line of reasoning I think is this:
Let $G$ be a finite subgroup of the diffeomorphism group of a surface $\Sigma_g$. Let $N(G)$ be the normalizer of $G$ in $Diff(\Sigma_g)$. So in most situations there's a short exact sequence $0 \to G \to \pi_0 N(G) \to \pi_0 Diff(\Sigma_g / G) \to 0$ (Birman-Hilden) where we're thinking of $\Sigma_g /G$ as an orbifold. Further there's a map $\pi_0 N(G) \to \pi_0 Diff(\Sigma_g)$. Provided the image of this map is the normalizer of $G$ in $\pi_0 Diff(\Sigma_g)$ (which according to Birman-Hilden is a pretty generic list of cases), then I think the argument goes through.
The Birman-Hilden reference. The case you're interested in is when $\Sigma_g / G$ is an orbifold whose underlying topological space is a sphere.
This is just off the top of my head and I'm likely glossing over some important details. I'll be happy to try and clarify.
So in general the lift of your moduli space is disconnected and its components are indexed by the cosets of the group $\pi_0 N(G)$ in the mapping class group. $\pi_0 N(G)$ is the hyperelliptic group in the case that $G = \mathbb Z_2$ is the hyperelliptic involution of your surface.