I think the assertion is always true. Here is my attempt at a proof - there should be an easier way to phrase this, but I don't seem to have a good feeling for filtrations, so it may look like overkill...
In the question and below, all functors are derived (including tensor products). Also $f^\ast$ just means the ordinary (derived) pullback of $D$-modules (which agrees with the pullback of the underlying $\mathcal O$-modules).
First note that it is enough to show that, for any $i$ and any element $u \in H^i f^\ast M$ the singular support of $D_X u$ is contained in $f_d f_\pi ^{-1} SS(M)$.
There is a natural extension of the $D$-module inverse image functor $f^\ast$ to a functor on graded modules for the Rees algebra $D_\hbar$, which has the expected behaviour at $\hbar=0$ (for example, see Section 3 of http://arxiv.org/abs/1603.07402).
In particular, if we pick a good filtration of the $D_Y$-module $M$, and consider the corresponding $D_{Y,\hbar}$-module $M_\hbar$, then there is a complex of (graded) $D_{X,\hbar}$-modules $K_\hbar = f^\ast M_\hbar$ such that
$$K_\hbar \otimes_{\mathbb C[\hbar]} \mathbb C_1 = f^\ast M,$$
and
$$K_\hbar \otimes_{\mathbb C[\hbar]} \mathbb C_0 = f_{d\ast} f_\pi^\ast (gr \ M).$$
Note that $H^iK_\hbar$ may have $\hbar$-torsion (or equivalently, may not be flat over $\mathbb C[\bar]$) so can't really be thought of as a filtered $D_X$-module in the usual sense. This was a bit confusing, so I tried to be careful about it.
Consider the $D_{X,\hbar}$-\module $H^i (K_\hbar)_{tf}$, given by the quotient of $H^i(K_\hbar)$ by the $\hbar$-torsion elements.
Note that $H^i (K_\hbar)_{tf}/\hbar$ is a quotient of $H^i (K_\hbar) /\hbar$, which in turn is a subquotient of $H^i(K_\hbar \otimes_{\mathbb C[\hbar]} \mathbb C_0 )$, which is a $\mathcal O_{T^\ast X}$-module supported in $f_d f_\pi^{-1} SS(M)$. Thus $H^i (K_\hbar)_{tf}/\hbar$ is itself supported in $f_d f_\pi^{-1} SS(M)$.
On the other hand, $H^i(K_\hbar)_{tf}/(\hbar-1) \simeq H^i(f^\ast M)$ (everything is flat over $\mathbb C[\hbar]$ when $\hbar \neq 0$). In other words, $H^i(K_\hbar)_{tf}$ defines a compatible filtration on the (possibly non-coherent) $D_X$-module $H^i f^\ast M$.
Now, given $u\in H^i(f^\ast M)$, we can lift it to an element $u_\hbar \in H^i(K_\hbar)_{tf}$, and consider $D_{X,\hbar} u_\hbar \subseteq H^i(K_\hbar)_{tf}$. This is a finitely generated $D_{X,\hbar}$-module which specializes to $D_X u$ at $\hbar=1$ (in other words, it defines a good filtration of $D_X u$). But the associated graded $D_{X,\hbar}u_\hbar /\hbar$ is a submodule of $H^i(K_\hbar)_{tf}/\hbar$ which is supported in $f_d f_\pi ^{-1} SS(M)$. Thus $SS(D_X u) \subseteq f_d f_\pi ^{-1} SS(M)$, as required.
