As Ben wrote, there are examples of non-simply connected strata.
For a quiver variety corresponding to a fundamental representation, it, a priori, only has a small group action. So it is usually far away from homogeneous.
So I suspect the answer is NO, though I do not know a concrete example.
But, I think the question itself is not right:
It is true that perverse sheaves appearing in the push-forward from `natural resolutions' are IC sheaves associated with constant sheaves. But it is not because stratum are simply-connected, or equivariantly simply connected. It is from a very different reason:
This reason cannot be seen if one treat an individual quiver variety separately. One can see it only when one consider various quiver varieties simultaneously, and relate them to a representation. It is basically because there is a component consisting of a single point, and other components are `connected' to it in some sense.
(Each component corresponds only to a weight space, and considering several components simultaneously, one get the structure of a representation. The single point corresponds to the highest weight vector.)
In summary, quiver varieties are not homogeneous in a conventional sense, but have substitutive property (I do not know how to call it) when one treat several components simultaneously.
That is way too ambitious, I think, considering what is known about classification of algebraic varieties. Ignoring the trivial, for these purposes, case of curves, the next and best studied case is algebraic surfaces. For them, the classification is classical and has been known for almost a century. So that is a good first case to consider.
So you are asking: what are the smooth projective surfaces $X$ with $\pi_1(X)=1$, and in particular with regularity $q=h^1(\mathcal O_X)=0$. Well, pick up your favorite text, Shafarevich etc. or Barth-(Hulek-)-Peters-van de Ven, or whatever, and go through the list.
Kodaira dimension $-\infty$: here you get all rational surfaces, and only these.
Kodaira dimension $0$: K3 surfaces.
Kodaira dimension $1$, elliptic surfaces $X\to C$ (a general fiber is elliptic). Clearly, $C$ must be $\mathbb P^1$. But actually getting $\pi_1(X)=1$ seems like not a completely trivial condition, something to think about.
Kodaira dimension $2$, i.e. surfaces of general type. Well, the examples of simply connected surfaces of general type are highly prized, especially if they have $p_g=h^0(\omega_X)=0$. Many such surfaces are known (e.g. http://en.wikipedia.org/wiki/Barlow_surface, some Godeaux surfaces, some Campedelli surfaces) but a complete classification? Not even close. Like I said, this is too ambitious for the present state of knowledge.
Best Answer
Yes! (I assume it was implicit in your question that the variety be projective?)
More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book ("Higher dimensional algebraic geometry").
Alternatively, take a look at Debarre's Bourbaki talk ("Varietes rationnellement connexes"). The idea is that a rationally connected variety has no holomorphic forms, so by Hodge theory the structure sheaf $\mathcal{O}_X$ is acyclic, implying $\chi(X,\mathcal{O}_X) = 1$. If $f : Y \to X$ is a connected etale cover, then $Y$ is again rationally connected, so by the same argument $\chi(Y,\mathcal{O}_Y) = 1$, so $\deg{f} = 1$. So $X$ is simply connected.
In positive characteristic the Hodge theory fails, so the argument as such doesn't stand, but you may still get the simple connectedness as a consequence of the fibration theorem of Graber-Harris-Starr-de Jong. See the mentioned Bourbaki expose.