electromagnetism - Mass-Energy equivalence in case of minimal coupling

The energy-momentum relation of a free particle is (in SI Units):

$$m^2c^4 =- c^2 \vec{p}^2 + E^2$$ Minimal coupling is a way to fix a gauge freedom for the choice of canonical momentum (which I can in special relativity give as $p_{\mu} = \left( \begin{matrix} \frac{E}{c} \\ p_x \\ p_y \\ p_z \end{matrix} \right)$. One way to write the minimal coupling would be by writing the canonical momentum now as:

$$p_{\mu} = \left( \begin{matrix} \frac{E - e \Phi}{c} \\ p_x + A_x \\ p_y + A_x \\ p_z + A_z \end{matrix} \right)$$ (Here I emply kind of a relativistic hamiltonian formalism, where the movement of a particle in space time, parameterised by $s$, is given by the hamiltonian $\lambda (m^2 + p^{\mu} p_{\mu})$, with $\lambda$ being an abitrary positive function of s, that describes how fast the path in space time is gone through).

Does this mean that I can generalize the energy-momentum relation to:

$$m^2c^4 = - c^2 (\vec{p} + \vec{A})^2 + (E-e\Phi)^2~?$$

If I solve this for $E$ for $\vec{A} = 0$, it resolves to:

$$E = e \Phi + \sqrt{m^2 c^4 + c^2 p^2}$$ which to me seems plausible, and that's why I'm asking.

Edit: the question is not only wether the modified relation holds, but also wether the reasoning behind is right.