Let $G = \{G_1 \rightrightarrows G_0\}$ be a finite groupoid, i.e. $G_1,G_0$ are both finite sets, and let $A$ be $\mathbb Q$-module. Regard $A$ as a discrete groupoid $A \rightrightarrows A$, and let $f: G\to A$ be a functor — equivalently, there is a set $G_0/G_1$ of isomorphism classes of $G$, and $f$ is an $A$-valued function on this set.
Then Baez and Dolan define a **groupoid integral**:
$$ \int_G f = \sum_{x\in G_0/G_1} \frac{f(x)}{\lvert {\rm Aut}(x)\rvert} $$
where to make the above precise I'm using the fact that if $x,y \in G_0$ are isomorphic, then $\lvert {\rm Aut}(x)\rvert = \lvert {\rm Aut}(y)\rvert$, and choosing an isomorphism $x \to y$ in fact induces an isomorphism ${\rm Aut}(x) \to {\rm Aut}(y)$. Anyway, the point is that $\int_G f$ actually depends only on the "stack" $G_0 // G_1$, where for our purposes "stack" can mean "groupoid up to equivalence": if $G,G'$ are equivalent groupoids and $f':G' \to A$ is the functor corresponding to $f$ under the equivalence, then $\int_Gf = \int_{G'}f'$.

Suppose now that $A$ is not a $\mathbb Q$-module but some (possibly weak) version of a "$\mathbb Q$-module in groupoids". (If you want, I have no objection to you thinking of $A$ as a strict object — the weakened version would replace all the axioms for a $\mathbb Q$ vector space with natural isomorphisms that have their own coherency.) Then it still makes sense to talk about functors $f: G \to A$.

Question:Does there exists an extension of the groupoid integral above to integrals of the form $\int_Gf$ where $f: G \to A$ is a functor but $A$ is not discrete, and that plays well with equivalences of groupoids $G \to G'$ and $A \to A'$?