Suppose that $\pi:E \to D$ is a 4-dimensional Lefschetz fibration over a disk,(more general, Lefschetz fibration over a surface with boundary ) and let $\Omega$ be a closed 2-form on $E$ such that it is non-degenerate fiberwise.

For any $x \in E$, there is a decomposition $TE_x=TE_x^h \oplus TE_x^v$, $TE^v=\ker d\pi$ and $TE_x^h=\{v \in TE_x: \Omega(v,w)=0,\forall w \in TE_x^v\}$. Let $\partial_t$ be the coordinate vector field on $S^1=\partial D$, $R$ be the horizontal lift of $\partial_t$ and $\phi_{\Omega}$ be the flow generated by $R$. Taking base point $1 \in S^1$, then we get a map $\phi_{\Omega}: \pi^{-1}(1)\to\pi^{-1}(1)$, so the boundary $Y=\pi^{-1}(\partial D)$ is a mapping torus $Y_{\phi_{\Omega}}$.

My Question: Suppose that $f: \pi^{-1}(1)\to\pi^{-1}(1)$ is a Hamiltonian perturbation of $\phi_{\Omega}$, can we find a symplectic form $\Omega'$ on $\pi:E \to D$ such that $\phi_{\Omega'}=f$ ?

I tried to reason as follows, but I am not sure.

Let $\omega$ be restriction of $\Omega$ on $Y=\pi^{-1}(\partial D)$ and $\lambda=\pi^*(dt)$, then $(\omega,\lambda)$ form a stable Hamiltonian structure on $Y_{\phi_{\Omega}}$. In a neighborhood of $\partial E$, we can identify $(E,\Omega)$ with $(Y\times[0,-\epsilon), \omega+ds\wedge\lambda)$.

Suppose that $f: \pi^{-1}(1)\to\pi^{-1}(1)$ is a Hamiltonian perturbation of $\phi_{\Omega}$, we can define a mapping tours $Y_f$. Let $\omega_f$ be unique closed two form on $Y_f$ such that $\omega_f=\omega$ on fiber. Since $f$ Hamiltonian isotopic to $\phi_{\Omega}$, so $[\omega_f]=[\omega] \in H^2(Y ,\mathbb{R})$.

Finding a family of close two form $\omega_s$ on $Y$ such that $\omega_s=\omega_f$ near $s=0$ and $\omega_s=\omega$ near $s=-\epsilon$. Replacing $(Y\times[0,-\epsilon), \omega+ds\wedge\lambda)$ by $(Y\times[0,-\epsilon), \omega_s+ds\wedge\lambda)$, then we obtain a new symplectic form $\Omega'$ on $E$ and I hope that $\phi_{\Omega'}=f$.