A bifibration is presumably equivalent to the data of a functor from T to categories, with the property that the map between categories induced by a map between spaces has an adjoint. So we can understand your bifibration D -> T as a contravariant functor that takes a topological space to its category of sheaves, and a continuous map to the pullback operation.

If by "category of sheaves" we mean the 1960's style triangulated category, this functor has bad properties--e.g. there is no simple way to recover D(X) from the categories D(U), where U runs through an open cover of X. But by now there is better technology: we can replace the triangulated category of sheaves by a suitable infinity-category. In this setting, D(X) will be an inverse limit in the infinity-categorical sense of the categories D(U) belonging to an open cover. This is a nice way of formulating what is sometimes called cohomological descent. We actually have descent for more general kinds of covers, for instance coming from free group actions. If Y is a principal G-bundle over X, then D(X) is the inverse limit of the cosimplicial diagram of infinity-categories D(Y x G x ... x G). This is the kind of thing that makes the equivariant theory of sheaves work.

So we can try to answer your question like this. A candidate theory of equivariant sheaves is a contravariant functor D_G from T^G to infinity-categories. It must agree with the usual theory of sheaves on T, which we identify with the full subcategory of T^G consisting of free G-spaces. And it must satisfy a strong enough kind of descent. After pinning down that last condition, I am pretty sure this will uniquely determine D_G.

You could translate all this back into the language of bifibrations (well, infinity bifibrations), but that point of view isn't so easy for me.

(Another way to put this is that D is itself a sheaf of categories on T in a suitable Grothendieck topology, and that this extends in a unique way to sheaf of categories on T^G. But I'm worried that there is some technical problem with this statement.)