Is it true that every smooth rational variety X is simply connected? How is the proof?
Would it be still true if X has mild (for example orbifold) singularities?



Are rational varieties simply connected?
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.


