$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$.

$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$.

EDIT This ad hoc answer can be extended as follows: All I really used was that $\pi_2(G)$ and $\pi_1(G)$ are finitely generated. But $\pi_n(G)$ is finitely generated for all $n\ge 1$ (reduce to simply connected case and use homology), so $\pi_n(G/H)$ is finitely generated for $n\ge 2$. That leads to a lot more higher-dimensional non simply connected examples.