(To match the standard definitions, you want $f$ or at least $df$ to have compact support. This doesn't really matter for this simple computation. I just need the Hamiltonian flow to be complete.)

Let $\gamma \colon [0,1] \to \mathbb{R}^2$ be a smooth path. Then, define a map
$\bar \gamma \colon [0,1]\times[0,1] \to \mathbb{R}^2$ by $\bar\gamma(s,t) = \Phi_t( \gamma(s))$, where $\Phi_t$ is the time $t$ flow of the Hamiltonian vector field of $f$. The area swept out by $\bar \gamma$ is obtained by integrating
\[
\int \bar\gamma^*\omega = \int \omega( T\Phi_t(\gamma(s))\cdot \dot \gamma(s),
X_f( \Phi_t (\gamma(s))) ds\wedge dt = \int_0^1 \int_0^1 df(\dot \gamma) ds dt = \int_0^1 df( \dot \gamma(s)) ds = f(\gamma(1)) - f(\gamma(0)).
\]
(I have taken the convention that $\omega(X_f, \cdot) = - df \cdot$. This matches your sign convention also. In simplifying this, I have used the time-independence of $f$.)

Take $p$ and $q$ to be maximum and minimum of the function $f$ and $\gamma$ a path connecting them that is to "one side" of the disk (but very close). If the time 1 map of $\Phi$ displaces the disk, it will move the path to the "other side" of the disk. The points $p$ and $q$ are kept fixed by the flow, since they are critical points, so the image of the rectangle is now a region that contains the disk. Its area is thus at least $\pi$, but the above computation shows this is the difference between max and min.

There are clearly some points in here that could use a few extra details to nail the ideas down completely -- in particular, this "one side" and the "other side" are compelling in pictures, but could use a little more care.