This is very rough and probably not too precise but here is what I
think.
For a two-form $\omega$ on a manifold $M$ denote by
$\text{null}(\omega_x)$ the vector subspace of the tangent space
$T_xM$ defined as follows:
$$\text{null}(\omega_x) = \left\{ \, W \in T_xM \,\, | \,\, \omega_x(W,v) = 0
\, \text{ for all } \, v \in T_xM \, \right\},$$ i.e. the
null-space of $\omega$ contains all tangent vectors that are
orthogonal to the whole tangent space $T_xM$. In the case when
$M=E$ and the assumption that $\Omega$ is non-degenerate
fiber-wise, $\dim{\text{null}(\Omega)} = 2$ or $0$. When the
dimension of the null-space is $0$ then $\Omega$ is non-degenerate
on $E$ and hence it is symplectic. Otherwise, it is symplectic on
the fibers only. However, when restricted to the even-dimensional
(more precisely three-dimensional) fibration over the unit circle,
the null-space is one-dimensional in both cases, so it defines a
line field.
Let $z = y_1 + i y_2$ be a notation for the points on the disc $D
= \{ z \in \mathbb{C} \, : \, |z| < 1 + \varepsilon \}$. In polar
coordinates we have $z = R e^{i\theta}$ where $R = |z|$ and
$\theta = \text{Arg}(z) \in [0,2\pi)$. Define the functions
$R^2(x) = |\pi(x)|^2$ and $\theta(x) = \text{Arg}(\pi(x))$. For $z
\in D$ let $E_z = \pi^{-1}(z)$ be the fiber over $z$. Now, the
lift of the standard symplectic form $2 \, dy_1 \wedge dy_2$ can be
written as
$$\pi^*(2 \, dy_1 \wedge dy_2) = d (R^2) \wedge d \theta.$$ Denote by $M_1 = \{ x \in E \, : \, |\pi(x)|^2 = 1 \}$. Take
$$l(x) = \{ W \in T_xM_1 \, : \, \Omega_x(W,v) = 0 \, \text{ for
all } \, v \in T_x E_{\pi(x)} \, \}.$$ This is a line field and
is equal to $T_xE^h \cap T_xM_1 = l(x)$, i.e. $l(x)$ is a one
dimensional line field tangent to $M_1$ and $\Omega$-orthogonal to
the fibers. Observe that you can also write it as the null-space
of $\omega_x = \Omega_x - d R^2 \wedge d \theta$. The lift of the
vector field tangent to the unit circle in the $D$ lies on $l(x)$.
Take any function $H : E \to \mathbb{R}$ and restrict it to $M_1$.
If you look at the form $\Omega'_x = \Omega_x - d(R^2) \wedge
\theta - d H \wedge d \theta$ restricted to $M_1$, then take
its one dimensional null-space $l_H(x)$. It gives
you a line field on $M_1$. The vector field on the unit circle
lifts to a vector field whose span defines $L_H(x)$. Furthermore,
its tangent trajectories give rise to a large family of
symplecotmorphisms (parametrized by $H$) on the fiber
$\pi^{-1}(1)$ by using the constriction of first return-map. Then,
using the null-space of $\omega$, you can trivialize $M_1$ by
creating the cover $p : E_1 \times \mathbb{R} \to M_1$ and as a
result you obtain more or less something like a time-dependent
(actually, $\theta$-dependent) Hamiltonian $\tilde{H}(y,\theta)$
where $(y,\theta) \in E_1 \times \mathbb{R}$ that is
$2\pi-$periodic and gives rise to a symplectomorphism on $E_1$ (a
version of the Poincare first return map). Then it looks like
generically, it is possible that its periodic points are isolated.
So a deformation of $\Omega$ that gives rise to non-degenerate
periodic orbits can be obtained by adding $d\theta \wedge \big(R^2
+ H \big)$. Observe that although $\theta$ is not well-defined on
the unit circle, its differential $d \theta$ is, because $d \theta
= d(\theta + 2\pi)$. Furthermore, if you feel uncomfortable about
$d \theta$ around the center of $D$, then multiply your whole
perturbation $d\theta \wedge \big(R^2 + H \big)$ by an infinitely
smooth bump-function which is constantly $1$ everywhere in $D$
except for a very small disc neighborhood around its center and it
is zero at the center.
