Not an answer yet, but some thoughts that may lead to one...
Let $$G(x) = \int \mathbb{1}_{\pi(y)\geq \pi(x)} ~dy$$
then $$I = \int \int_{x,y} \min( \pi(x), \pi(y) )~dx~dy = 2 \int \pi(x) G(x)~dx$$
A sufficient condition would be to that show $G(x) = \mathcal{O}(||x||^d)$. This is satisfied if $\pi$ is spherical, representing the decay of the tail. If the distribution is not radial, it could have a long, flat, thin ridge that extends to infinity, upon which $G$ remains constant. It's clearly not a necessary condition.
edit: seems there's already an elegant answer