Prove that $E_\pi(\Phi(X))=E_q(\Phi(X)\frac{f_\pi(X)}{f_q(X)})$ when $\pi<< q$, and both $q,\pi$ are absolutely continuous with respect to Lebesgue measure $m$.
$$E_\pi(\Phi(X))=\int_\Omega\Phi(X)d\pi=\int_\Omega\Phi(X)\frac{d\pi}{dq}dq=\int_\Omega\Phi(X)\frac{d\pi}{dm}\frac{dm}{dq}dq=\int_\Omega\Phi(X)\frac{d\pi/dm}{dq/dm}dq=E_q(\Phi(X)\frac{f_\pi(X)}{f_q(X)})$$
The step of inverting $$\int_\Omega\Phi(X)\frac{d\pi}{dm}\frac{dm}{dq}dq=\int_\Omega\Phi(X)\frac{d\pi/dm}{dq/dm}dq$$ requires $q$ equivalent to Lebesgue measure $m$. But this may be too strong thus fails for compactly supported distributions.
When you calculate an expectation w.r.t. $q$ you need only integrate over $\{x:f_q(x) \neq 0\}$ Restricted to this set $q$ is equivalent to Lebesgue measure and your computation is valid. Note that $\frac {f_{\pi}} {f_q}$ is not even defined at all points. So you have to interpret that right side of your equation appropriately by restricting your attention to $\{x:f_q(x) \neq 0\}$. Once you do this the formula is valid even when $f_q$ has compact support.