Let $(M,\omega)$ be a symplectic manifold and let $X\colon M\rightarrow TM$ be a vector field on $M$.
Theorem. The vector field $X$ is Hamiltonian if and only if the following two conditions are fulfilled:
If $(\phi_t)_t$ is the flow of $X$, one has ${\phi_t}^*\omega=\omega$,
For all loop $c\colon]-1,1[\rightarrow M$, one has $\displaystyle\int_ci_X\omega=0$.
Proof. Let us proceed by double implication.
Assume $X$ is Hamiltonian, then there exists a smooth map $H\colon M\rightarrow\mathbb{R}$ such that: $$i_X\omega=\mathrm{d}H.$$ Therefore, the differential form $i_x\omega$ of degree $1$ is exact and its integral vanish on all loops drawn on $M$. Furthermore, using Cartan's formula for the Lie derivative and since $\omega$ is closed, one gets: $$\frac{\mathrm{d}}{\mathrm{d}t}({\phi_t}^*\omega)_{\vert t=0}=0.$$ I do not know how to deduce that ${\phi_t}^*\omega=\omega$. If I use directly the definition of the pullback, I get: $$({\phi_t}^*\omega)_x(v,w)=\omega_{\phi_t(x)}(\mathrm{d}_x\phi_t\cdot v,\mathrm{d}_x\phi_t\cdot w),$$ which does not look to promising.
I will assume that $M$ is connected (is it necessary?). Assume that $X$ fulfils the two given conditions, let $m\in M$, for all $x\in M$, I would like to define: $$H(x):=\int_{c_x}i_X\omega,$$ where $c_x$ is any path from $m$ to $x$. According to the second condition, the result will not depend on $c_x$. Therefore, the only difficulty is to choose it so that $H$ is smooth and I don't know how to proceed.
Whence the result. $\Box$
Any enlightenment will be greatly appreciated.
$$\frac{\mathrm{d}}{\mathrm{d}t}({\phi_t}^*\omega)_{\vert t=t_0}=\phi_{t_0}^*(\frac{\mathrm{d}}{\mathrm{d}t}({\phi_t}^*\omega)_{\vert t=0}).$$ implies that $\phi_t^*(\omega)$ is constant and is equal to $\phi_{0}^*(\omega)=\omega$ since $\phi_0^*$ is the identity.