That's because symplectic cohomology field theory is secretly an $S^1$-equivariant theory. As a finite-dimensional model, suppose that you have a manifold $M$ with an $S^1$-action and an invariant function $h: M \rightarrow \mathbb{R}$. Take an orbit of the $S^1$-action (not a fixed point) which consists of critical points of $h$, and such that the Hessian is transversally nondegenerate. If the negative eigenspaces of the Hessian form a nontrivial vector bundle over our orbit, the local contribution to the Morse homology is a twisted homology of $S^1$, which vanishes with rational coefficients. This vanishing also holds for equivariant homology. Note that this phenomenon can never happen for free orbits, since the $S^1$-action itself provides a trivialization of the bundle of negative eigenspaces, but it does happen for orbits with finite even stabilizer.
The analogy is of course that $h$ corresponds to the action functional on free loop space. Free orbits of critical points correspond to simple periodic (Reeb) orbits, and ones with finite stabilizers to multiple covers of simple orbits.
That's because symplectic cohomology is secretly an $S^1$-equivariant theory. As a finite-dimensional model, suppose that you have a manifold $M$ with an $S^1$-action and an invariant function $h: M \rightarrow \mathbb{R}$. Take an orbit of the $S^1$-action (not a fixed point) which consists of critical points of $h$, and such that the Hessian is transversally nondegenerate. If the negative eigenspaces of the Hessian form a nontrivial vector bundle over our orbit, the local contribution to the Morse homology is a twisted homology of $S^1$, which vanishes with rational coefficients. This can never happen for free orbits, since the $S^1$-action itself provides a trivialization of the bundle of negative eigenspaces, but it does happen for orbits with finite even stabilizer.
The analogy is of course that $h$ corresponds to the action functional on free loop space.