Work of Gan and Ginzburg (https://arxiv.org/pdf/math/0409262v7.pdf) shows that one of our favorite examples, $M=T^*(\mathfrak{gl}_n\oplus \mathbb C^n)$ and $G=GL(n)$, has a 0 level of the moment map which has $n+1$ components and so is not irreducible (but for a given stability condition, all but one component is unstable).

**EDIT:** I'll just add that this situation can get quite bad; in the example above, at least things are equidimensional. Even this can fail, and $\Phi^{-1}(0)$ can have a larger dimension than you would expect from the GIT quotient, since only a component of smaller dimension has stable points; this happens for all quiver varieties corresponding to non-dominant weight spaces. For example, if $M=T^*\mathrm{Hom}(\mathbb C^n,\mathbb C^m)$ and $G=GL(n)$, then we have a pair of matrices $A$ and $B$ subject to $BA=0$ as the moment map condition. Obviously, the ranks of these matrices can sum to no more than $m$ or to $2n$. If $2n\leq m$, then there's a single component.
If $m<2n$ (which is the non-dominant case; this is a $\mathfrak{sl}_2$ quiver variety), then there are $m-2n+1$ components, given by the closures of the sets where $A$ has rank $k$ and $B$ rank $m-k$. The dimension of this space is given by choosing a dimension $k$ subspace in $\mathbb C^m$, the surjective map of $\mathbb C^n$ to this space, and injective map from its quotient. This gives dimension $\binom mk+nm$. Thus, the largest component is when $k\approx m/2$, but the two different stability conditions are that $A$ is injective or that $B$ is surjective (so $k=n, m-n$); thus, if $n>m/2+1$, we're no longer equidimensional.

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