Answer to part two: No - this cannot be - even if you relax the conditions and only assume the vector field to be smooth.

Indeed, the question can be generalized somewhat: let $X : N \to TN$ be any smooth vector field on a manifold $N$. Then we may define a Hamiltonian

$H : T^*N \to \mathbb{R} \qquad$ by $\qquad H(q,p)=p(X_q)$.

Here $T^*N$ denotes the cotangent bundle (which has a canonical symplectic structure), and hence it makes sense to evaluate a cotangent vector $p \in T^*_qN$ on a vector in $T_qN$. This generalizes your second question with $X(x,y)=(Q(x,y),P(x,y))$ and $N=\mathbb{R}^2$.

The flow of the associated Hamiltonian vector field $X_H$ has flow $\Phi_t$ the same as the flow $\varphi_t$ of $X$ but lifted to the unique symplectomorphism, which is linear in $T_q^*N \to T^*_{\varphi(q)}N$. This is also given as:

$\Phi_t(q,p)=(\varphi_t(q),p\circ (D_q\varphi_t)^{-1})$

Indeed, if you (in local coordinates on $N$ - so we are now back in $\mathbb{R}^{2n}=T^*\mathbb{R}^n$) differentiate this w.r. to $t$ at $t=0$ you get the vector field:

$(X_q,-p \circ (\nabla X)_q)$

This is (up to a sign) precisely equal to $J_0\nabla H$.

Consequences of this are: Any periodic orbit (i.e. a fixed point $(q,p)=\Phi_T(q,p)$ must project to a similar orbit for $X$, and have: $p=p \circ (D_q\varphi_T)^{-1}$. This means that any eigenvector with eigenvalue 1 for $D_q \varphi_T$ defines such an orbit - so if 1 is an eigenvalue there are infinitely many. Since this return map takes any vector tangent to the orbit to itself we have such an eigenvector.

About the first question: I doubt it: indeed, no orbit can be non-degenerate since these are stable (for fixed orbit time $T_0$) under pertubations - hence there must be one with period $T$ for each $T\in[T_0-\epsilon,T_0+\epsilon]$ close by (for small enough $\epsilon>0$).