I'll show you how to do this in SI units, instead of CGS. The question is to derive the Lorentz force law using Hamilton's equations and the Hamiltonian of a charged particle in the presence the vector and scalar potentials, ie.
$$H=\frac{(\vec p-e \vec A)^2}{2m}+e V ,\;\;\;\;\;\;(1)$$
where e and m are the charge and mass of the particle respectivley. Hamilton's equations are
$$\dot q_i=\frac{\partial H}{\partial p_i} \;\;\;\;\;\;(2)$$
and
$$\dot p_i=-\frac{\partial H}{\partial q_i} , \;\;\;\;\;\;(3)$$
where the dot refers to ordinary differentiation with respect to time. In order to follow the solution, you must be familiar with Einstein summation convention, the Levi-Civita symbol and the Kronecker delta
Substituting eq'n (1) into (2) gives
\begin{align*}\dot q_i =& \frac{\partial}{\partial p_i}\left(\frac{1}{2m}\sum_{j=1}^3\left(p_j-e A_j\right)^2\right) \\
=&\frac{1}{m}\left(p_i-e A_i \right) .\end{align*}
Multiplying both sides by m gives
$$m \dot q_i = p_i -e A_i \;\;\;\;\;\;(4)$$
Differentiating with respect to time yields,
$$m \ddot q_i = \dot p_i -e \dot A_i \;\;\;\;\;\;(5)$$
Now substituting in eqn (1) into (3) and keeping in mind that the vector potential A is a function of the coordinates,
\begin{align*}\dot p_i=\frac{e}{m}\sum_j \left(p_j - e A_j\right)\frac{\partial A_j}{\partial q_i} -e \frac{\partial V}{\partial q_i} \end{align*}
and substituting in eqn (4) gives
$$\dot p_i=e v_j \partial_i A_j - e \partial_i V \;\;\;\;\;\;(6)$$
where I'm using Einstein summation convention (summation of the index j here) and I've defined
$$v_j\equiv\dot q_j$$
and
$$\partial_i\equiv\frac{\partial}{\partial q_i} .$$
Plug (6) into (5) to yield,
$$m \ddot q_i =e v_j \partial_i A_j - e \partial_i V - e \dot A_i\;\;\;\;\;\;(7)$$
(note we're still using Einstein summation convention)
Now,
\begin{align}\frac{d A_i}{dt}=&\frac{\partial A_i}{\partial t}+\frac{\partial A_i}{\partial q_j}\frac{dq_i}{dt} \\
\dot A_i =&\frac{\partial A_i}{\partial t}+v_j \partial_j A_i\end{align}
Substituting this result into (7) and rearranging a bit, we get
$$m \ddot q_i = e \left(v_j \partial_i A_j - v_j \partial_j A_i \right)+e\left(-\frac{\partial V}{\partial q_i}-\frac{\partial A_i}{\partial t}\right) \;\;\;\;\;\;(8)$$
I need to prove that the first term here is actually just the i-th component of v cross B, so to do that I employ the Levi-Civita symbol,
\begin{align}\left(\vec v \times \vec B\right)_i = \left(\vec v \times\left(\vec\nabla \times \vec A\right)\right)_i=&\epsilon_{ijk}v_j \left(\vec\nabla \times \vec A\right)_k \\
=&\epsilon_{ijk}v_j\epsilon_{klm}\partial_l A_m \\
=&\epsilon_{ijk}\epsilon_{lmk}v_j \partial_l A_m
\end{align}
and using the identity (see Wiki's article on Levi-Civita),
$$\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$,
we get
$$\left(\vec v \times \vec B\right)_j=v_m\partial_i A_m - v_j \partial_l A_i$$
or upon relabelling,
$$\left(\vec v \times \vec B\right)_j=v_j\partial_i A_j - v_j \partial_j A_i \;\;\;\;\;\;(9)$$
We'll use this in a second, but first consider that
$$\vec E = -\vec \nabla V - \frac{\partial \vec A}{\partial t}$$
(see Griffiths EM book) or in component form,
$$E_i = -\frac{\partial V}{\partial q_i} - \frac{\partial A_i}{\partial t} \;\;\;\;\;\;(10)$$
Finally, substituting (9) and (10) into (8) yields
$$m \ddot q_i = e\left(\vec v \times \vec B \right)_i + e E_i$$
or written in vector form,
$$m \ddot{\vec{q}} = e\left(\vec E +\vec v\times\vec B\right)$$
which is the Lorentz force law. QED.
