I) The Hamiltonian for point charges and EM fields can certainly produce the EOMs of the particle(s) as well as the EM fields.
A full explanation is quite a long story. For pedagogical reasons, to see how this works, it is best to:
Firstly, understand the corresponding Lagrangian formulation.
Secondly, understand how the Hamiltonian formulations work for point charges and EM fields separately, see e.g. this & this Phys.SE posts.
Thirdly, try to construct a Hamiltonian formulation for both point charges and EM fields together.
II) One correction: OP's Hamiltonian (1) yields the correct total energy, but OP asks how to produce Maxwell's equations. For the latter purpose, OP's Hamiltonian (1) is missing a Lagrange multiplier term that imposes Gauss' law.
III) Concretely, the minimal phase space is as follows:
Particle position ${\bf r}(t)$ and particle momentum ${\bf p}(t)$:
$$\{r^k(t), p_{\ell}(t)\}= \delta_{\ell}^k.$$
(Minus$^1$) the electric field ${\bf E}(x)$ is the canonical conjugate variable to the magnetic gauge potential ${\bf A}(x)$: $$\{A_i({\bf x},t), E^j({\bf x}^{\prime},t)\}~=~ -\delta_i^j~\delta^3({\bf x}\!-\!{\bf x}^{\prime}).$$
Lagrange multiplier $A^0(x)\equiv \phi(x)$.
IV) The equations come about as follows:
The magnetic field ${\bf B}\equiv{\bf \nabla}\times {\bf A}$ is defined as the curl of the magnetic gauge potential ${\bf A}$.
The Hamilton's equations for ${\bf r}$ and ${\bf p}$ yield (i) the
Newton's 2nd law with a Lorentz force, and (ii) the relation between velocity $\dot{\bf r}$ and momentum ${\bf p}$.
The Hamilton's equations for ${\bf A}$ and ${\bf E}$ yield (i) the
Maxwell–Ampere's law, and (ii) the relation between the electric field ${\bf E}$ and the gauge potential $A_{\mu}$.
The Lagrange multiplier $A^0\equiv\phi$ imposes Gauss' law.
The source-free Maxwell equations follows from the existence of the gauge potential $A_{\mu}$.
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$^1$ We use $(-,+,+,+)$ Minkowski sign convention with $c=1$.
