This is a question near and dear to my heart, since it is actually what my thesis will be about. Let me denote the Lie algebroid of a groupoid $G$ with base $M$ by $A$.
If you're viewing the stack presented by $G$ as a smooth model for the quotient space $M/G$, then the first proposal doesn't really seem to give the right answer. Namely, the Lie algebroid structure on $A$ induces a Poisson structure on $A^*$, and the quotient space of the groupoid $T^*G$ you mention is just the space of symplectic leaves for this Poisson structure --- this space doesn't give the cotangent bundle of $M/G$ and doesn't have a natural symplectic structure of its own. (Of course, these quotients aren't necessarily smooth, but even if they are you don't get the right answer.) Indeed, if you start with a trivial groupoid $M$, the space you get is just $M$.
Probably the underlying question you first have to answer is: in what sense is $T^*$ a functor? As Andre says, if all you care about is etale maps there's no problem. In general, you still get something interesting if you change the target category for the $T^*$ functor to the so-called symplectic category, where objects are symplectic manifolds, and a morphism $M \to N$ is a lagrangian submanifold of $\overline{M} \times N$, where the bar denotes changing the sign of the symplectic structure. (The usual caveat applies: this is not really a category since compositions are not always well-defined, but we'll just ignore this for now.) Composition is just the usual composition of relations.
Now to a smooth map $f: X \to Y$ you can assign the lagrangian submanifold
$T^*f := \{ (x, df_x^*\xi, f(x), \xi) \} \subseteq \overline{T^*X} \times T^*Y$,
where $df$ is the tangent map. This then gives a nice functor from the category of smooth manifolds to the symplectic category. In particular, if $f: X \to Y$ is a diffeomorphism (or even just etale), then $T^*f$ is in fact the graph of an actual map.
The upshot is that if you start with a groupoid, you can apply this functor to everything in sight to get a "groupoid" in the symplectic category. (This idea is due to Alan Weinstein. I'm calling this a "groupoid" since it is not clear in what sense this object is actually a groupoid. In particular, the symplectic category doesn't have fiber products, so the notion of a groupoid object isn't defined.) If your groupoid is etale, this recovers Andre's suggestion. So, you can instead try to formulate the notion of the cotangent stack of a stack in terms of this object. This post has gone on long enough, so I won't go into details here, but numerous examples suggest that this is indeed the right thing (or at least "a" right thing) to consider. In particular, from this point of view the "symplecticness" of cotangent stacks comes from the fact that they live in the symplectic category (this can be made more precise).
I should also say that this construction is somehow related to the groupoid structure on $T^*G$ you mentioned, but it contains more information that just that.
Edit: Here's one example illustrating some of the above. Say $G$ is a group acting smoothly on $M$. Then the action lifts to a Hamiltonian action on $T^*M$ with momentum map $\mu$. When the action is free, the resulting symplectic quotient $\mu^{-1}(0)/G$ is exactly the cotangent bundle of $M/G$. Even when the action is not free, it makes sense to call the stack $[\mu^{-1}(0)/G]$ the cotangent stack of $[M/G]$. The "groupoid" resulting by performing the above construction on the action groupoid $G \times M$ indeed encodes this quotient.
