The answer is definitely no in general. But what if we require some additional conditions?
Let $F:(M,\omega)\to(M,\omega)$ be a symplectomorphism, i.e. $F^*\omega=\omega$. We further assume $F$ is isotopic to $\mathrm{id}_{M}$ in $\mathrm{Diff}(M)$ or even in $\mathrm{Diff}(M,\omega)$.
Question: Is it necessary that $F$ must be a Hamiltonian isotopy? If not, is there any counter-example?
There are two ways to read your question since you use the phrase "Hamiltonian isotopy", rather than the related notion of "Hamiltonian symplectomorphism"-i.e. a symplectomorphism which Hamiltonian isotopic to the identity.
V1. Is a symplectic isotopy a Hamiltonian isotopy?
No, see Proposition 9.19 of "Introduction to symplectic topology" (Second edition).
V2. Is a symplectomorphism which is isotopic (or even symplectically isotopic) to the identity a Hamiltonian symplectomorphism?
The answer is also no. See see the symplectomorphism of $T^*{S^{1}} $constructed here: Showing that some symplectomorphism isn't Hamiltonian. It is symplectically isotopic to the identity by $F_{t}(x,\xi) = (x,\xi + t)$, but it is not a Hamiltonian symplectomorphism as is proven there.