This isn't really an answer to Ben's question, more like an attempt to isolate one possible way of thinking about how to make $HH_*$ very explicit.

Let’s first look at a variety $X$ (assumed affine for the discussion so I can avoid too many irrelevant technicalities). The $HH_*$ of $D$-modules is typically computed in two ways, which are related but not identical. Both are computing the self-Tor over $D \otimes D$ of the diagonal $D$-bimodule. One of them first computes the derived monoidal product of $D$-modules on $X \times X$—that is, the derived $\mathcal{O}$-module tensor product—and then computes the de Rham cohomology of that. Standard $D$-module identities (projection formula basically) imply that this is computing $HH_*$. This calculation immediately generalizes to stacks (it only takes a few lines, even) to identify the $HH_*$ with de Rham cohomology of the inertia stack.

There's also another approach. Again, first for varieties, that’s to resolve $D$ as a $D$-bimodule “all at once” and then compute the tensor product as $D\otimes D$-modules directly, i.e., tensor over $D\otimes D$).

For special cases, like when $X$ is a vector space, one can use the (noncommutative version of the) Koszul resolution of the diagonal. This is implicitly realizing $D(X)$ as a quadratic algebra with inhomogeneous relations; the method is pretty standard (I like Kapustin-Kuznetsov-Orlov or Polishchuk-Positselski as readable references, but there are many good ones).

More generally, there’s a standard trick that seems to appear in basically all the literature in the 80s on $HH_*$ of deformation quantizations (Brylinski, Brylinski-Getzler, Wodzicki, etc.; actually I don't think I've checked Feigin-Tsygan but I would guess it's similar?). That’s to filter the algebra $D$ (again, a choice here) and then use the spectral sequence to compute $HH_*$ of the filtered algebra. The associated graded will be some commutative thing (in the case of $D$-modules on $X$, functions on $T^*X$) and you get its $HH_*$ via HKR; then you just have to compute the $E_1$-differential, which you find (using the conical structure of the commutative algebra and some halfway explicit calculation) is the de Rham differential. This is nicely explained in Brylinski-Getzler in an explicitly conical setting. Finally then you need that the spectral sequence degenerates at $E_2$, and a standard way is to use local coordinates (say in the $C^\infty$ setting) or a formal geometry argument (in algebraic geometry; see Bezrukavnikov-Kaledin for the standard formalism in symplectic algebraic geometry) to reduce to knowing this for the Weyl algebra, where you just crank it out using the previous paragraph. Now this line of attack “evidently” works to identify $HH_*$ of a filtered deformation with de Rham cohomology of the underlying conical symplectic variety.

Note the difference: when applied to the category of $D$-modules, the first attack naturally gives de Rham cohomology of the base $X$, the second de Rham cohomology of the cotangent bundle itself $T^*X$. Of course that’s not an important difference in the result, but the point is that in calculating sort of “base then fiber” (or maybe “fiber then base”) vs. “all at once” the fiber contribution in the first method gets naturally integrated out so you're not explicitly seeing it in the answer.

So, this second approach should microlocalize well. The trick is to resolve the kernel of the identity endofunctor of $D$-modules on the stack in microlocal terms, in a reasonably explicit way. In $D$-module terms, this endofunctor is the $D$-module push forward along the diagonal $X/G \rightarrow X/G \times X/G$ of the “constant sheaf,” which lifts under the flat covering $X \rightarrow X/G$ to the push forward of $\mathcal{O}$ along $G \times X \rightarrow X \times X$ (via action, projection). So that, at least, is not mysterious. The tensor product of $D$-modules in either sense (i.e. over O, which gives a monoidal structure on the category of D-modules, or over D, which lands in vector spaces and is exactly what we want to compute) is computed (or even better, defined!) by descent from $X \times X$, so in principle we can calculate it “upstairs” on the flat cover.

OK, now the trick is to formulate that calculation in a nice invariant way! Which unfortunately I haven't done...