Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $xi_0\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1$, \phi_{\ast}\omega_0=\omega_1$, and d\phi{\ast}h_0=dh_1$.
For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1$, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and d\phi{\ast}h_0(x_1)=dh_1(x_1)$.
So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.
In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.
Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df$, and determines a unique $X_t$ for each smooth function $f$.
Finally the second condition becomes the following one only on $f$: $Y_t(f)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.
A solution for this equation could be constructed using the method of characterics, considering that $Y_t$ is non singular because such is $dH_t$.