The answer to your question is yes, given that the Hamiltonian perturbation indeed is sufficiently small. Conceptually the idea of the construction is better formulated as follows: fix the symplectic form $\Omega$ and instead deform the *fibration* by a smooth (in general non-symplectic) isotopy in order to obtain a different monodromy. (If you instead insist on fixing the fibration, the symplectic form will be deformed by the pull-back under this smooth isotopy.)

First, construct a standard symplectic neighbourhood of $F=\pi^{-1}(1) \subset (E,\Omega)$ being the product $$(F \times [-\epsilon,\epsilon]^2,\Omega|_F \oplus dx \wedge dy).$$
We can moreover use an identification in which all $F \times \{(x,0)\}$ are fibres $\pi^{-1}(z)$ for $z \in \partial D$ in the base. However, all fibres for $z \notin \partial D$ can only be assumed to be $C^\infty$-close to the symplectic surfaces $F \times \mathrm{pt}$ (equality for all fibres would imply flatness of the fibration).

Second, take a smooth isotopy which deforms $(u,(x,y))$ to $(u,(x,H_x(\phi^t_{H_t}(u))+y))$ , $u \in F$, for a smooth function $H_t \colon F \to [-\epsilon,\epsilon]$. It is readily checked that the monodromy around $\partial D$ is deformed by the corresponding Hamiltonian diffeomorphim $\phi^\epsilon_{H_t}$. Note that you have to interpolate this smooth isotopy to make it defined on all of $E$, and also do this in away so that the image of a fibre is still symplectic. The latter can be done, but I am a bit brief at this point. Also, you need to be careful concerning the support of $H_t$.