GIT quotient of variety with finite quotient singularities Let $X$ be a variety over $\mathbb{C}$ with finite quotient singularities, i.e. every point has a Zariski-open neighbourhood isomorphic to $U/H$ where $U$ is a smooth variety and $H$ is a finite group acting on it.
Now assume we have a connected reductive group $G$ acting on $X$ with finite stabilizers, let $\mathcal L$ be an ample $G$-linearized line bundle on $X$
and let $Y = X^{ss}/G$ be the corresponding GIT quotient. If necessary we may assume that $X^{ss} = X^s$.
Is it true that $Y$ again has finite quotient singularities?
If $X$ is already smooth, then this should be true (see Mumford's GIT, Chapter 8, §4). 
 A: Since I asked the question, I have come to realize that in my application, I know more about $X$ and the action of $G$. For this application, $X$ has finite quotient singularities as it is the coarse moduli space of a smooth DM stack and the action of $G$ comes from an action of the group on this stack in the sense Romagny's paper. Using this, I think I am able to show the desired conclusion. Let me sketch the proof here:
Proposition
 Let $\mathcal{M}$ be an orbifold, that is a smooth separated Deligne-Mumford stack with connected coarse moduli space and containing a non-empty open substack which is a scheme. Let the smooth group scheme $G$ act on $\mathcal{M}$ with finite, reduced stabilizers at geometric points. Then the quotient $\mathcal{M}/G$ is again a smooth DM stack.
Proof
Consider the frame bundle $\mathcal{F} = \text{Fr}\left( T_\mathcal{M} \right)$ of the tangent bundle $T_\mathcal{M}$ of $\mathcal{M}$. We know $\mathcal{F}$ is an algebraic space and for $n=\dim(\mathcal{M})$, the group $\text{GL}_n$ acts on $\mathcal{F}$ on the right by
$$   (v_1, \ldots, v_n) . (a_{ij})_{i,j=1}^n = (\sum_{i=1}^n a_{i1} v_i, \ldots, \sum_{i=1}^n a_{in} v_i) $$
 and we have that $\mathcal{M} = [\mathcal{F}/\text{GL}_n]$. On the other hand, the action of $G$ on $\mathcal{M}$ induces an action of $G$ on $\mathcal{F}$ by
$$   g . (v_1, \ldots, v_n) = (g_* v_1, \ldots, g_* v_n), $$
 where $g_*$ denotes the pushforward under the map $p \mapsto gp$ on $\mathcal{M}$. One shows that both actions are strict actions on the stack $\mathcal{F}$ and that they commute (as pushforward is $\mathbb{C}$-linear) and hence induce an action
$$G \times \text{GL}_n \times \mathcal{F} \to \mathcal{F}$$
with finite reduced stabilizers.
With these preparations done, we simply note that
$$\mathcal{M}/G =  [\mathcal{F}/\text{GL}_n]/G = (\mathcal{F}/\text{GL}_n)/G = \mathcal{F}/(\text{GL}_n \times G).$$
The second isomorphism is a consequence of Theorem 4.1  and Remark 2.4 of Romagny's paper. But now $\mathcal{F}$ is a smooth algebraic space and the action of $G \times \text{GL}_n$ has finite, reduced stabilizers at geometric points, so the quotient $\mathcal{F}/(\text{GL}_n \times G)$ is again a smooth Deligne-Mumford stack. qed
Hence instead of first taking the coarse moduli space $X$ of $\mathcal{M}$ and then performing the quotient, one can first take the quotient stack, which is smooth DM, and then take the coarse moduli space, which thus has finite quotient singularities. 
I don't know whether this can be used to answer the original question. In Proposition 2.8 of a paper by Vistoli it is shown that any normal complex variety with quotient singularities is the coarse moduli space of a smooth stack. However, to apply this to the question above one would need a way to lift the action of $G$ to a strict action on this stack.
