There are counter-examples, here is the simplest one:

Let $M$ be the cylinder $S^1\times \mathbb R$ with the symplectic form $ds \wedge dt$. Then the Hamiltonial $H=t$ defines an $S^1$-action on the cylinder.

There are counter-examples, hope they answer your question completely, just take any non-simply connected $G$ and consider its action on $T^*G$. The simplest case is:

Let $M$ be the cylinder $S^1\times \mathbb R$ with the symplectic form $ds \wedge dt$. Then the Hamiltonial $H=t$ defines an $S^1$-action on the cylinder.

One more counterexamle. Consider just the action of $SO(3)$ on its cotangent space. Clearly this action is Hamiltonian. Let us take the subgroup $S^1\subset SO(3)$ that represents the non-zero element of $\pi_1(SO(3))$. Obviously all the orbits of the action of this $S^1$ on $T^*(SO(3))$ will not be contractible.

So we see that in the case the Lie group is not simply-connected it always admits a "bad" action.