If $H$ is the fixed points of an involution of the $G$ the answer is known completely, and $H$ is "almost always" not simply connected. The most obvious example is $SO(n)\subset SL(n)$.
Suppose $G$ (complex) is simple and simply connected, $H$ is the fixed points of an involution of $G$, and $H$ is semisimple. Then the fundamental group of $H$ is $\mathbb Z/2\mathbb Z$ except in the following cases: $Sp(2n)\subset SL(2n)$, $Spin(n)\subset Spin(n+1)$ ($n\ge 3$), $Sp(2p)\times Sp(2q)\subset Sp(2(p+q))$, $F_4\subset E_6$, and $Spin(9)\subset F_4$. The proof is a fairly simple, case-free, root/weight lattice argument, along the lines of some of the preceding discussion. See http://www.ams.org/mathscinet-getitem?mr=2112326.
A closely related question is When is a finite dimensional real or complex Lie Group not a matrix group
It would be interesting, and probably not too difficult, to extend this to the case when $H$ is the fixed points of an automorphism of finite order, and is maximal.