Electromagnetic black hole?
Such a singularity would not occur, if you have no lower bound on the negative charge. For gravity, the singularity occurs because gravitational potential energy and relativistic kinetic energy both depend on the mass of the smaller object, which allows it to divide out when you solve for escape velocity. However, in this case, only the electromagnetic potential energy, not the kinetic energy, depends on the negative charge. This means the negative charge never gets divided out, and hence you can arbitrarily decrease the magnitude of this charge to make the escape velocity as low as you want.
However, if you fix the negative charge at some value $q_2$, it will be unable to escape when its electromagnetic potential energy equals its rest energy (up to a sign). So just set $\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} = mc^2$ (again, up to the correct sign) and solve for $r$.
To provide a consistent relativistic treatment of the problem, I will model the "foam-ball" by the Reissner-Nordström metric which represents the relativistic gravitational and electromagnetic field of a point of effective mass $M$ and charge $Q$. The reason why I am calling $M$ the effective mass is that even the pure electromagnetic field necessarily carries some energy and will thus look from afar as a gravitating mass of some mass $M_Q$. I will not discuss here what $M_Q$ should be and just leave $M$ as a free, in principle non-zero parameter. (The only influence on this analysis is that this allows for the existence of an event horizon.)
I will also use the term event horizon, which is the usual horizon known from general relativity from behind which no particle can escape, and the term "electromagnetic horizon", which would be a point from behind which a particle of certain charge is unable to escape.
The Reissner-Nordström metric (in geometric units) reads $$ds^2 = -\left(1 - \frac{2 M}{r} + \frac{Q^2}{r^2}\right) dt^2 + \frac{dr^2}{1 - 2M/r + Q^2/r^2} + r^2 d\Omega^2$$ This solution has an event horizon (horizon for any particle) at $1 - 2M/r + Q^2/r^2 = 0$ if $M^2>Q^2$.
The electromagnetic potential $A_\mu$ has only one non-zero component which is $A_t = -Q/r$. A charged test particle in this field obeys the Hamilton's equations generated by the Hamiltonian $$ H = \frac{1}{2} g^{\mu \nu} (\pi_\mu - e A_\mu)(\pi_\nu - e A_\nu)\,, $$ where $\pi_{\mu} = m u_\mu + eA_\mu$ is the canonical momentum conjugate to $x^\mu$. Both the field and the metric are static and we thus know that the total energy of the particle $\mathcal{E}\equiv-\pi_t$ will be an integral of motion. Particles at infinity have total energy larger than $m$, $\mathcal{E}>m$. I.e., a particle which has escaped the potential wells of the electromagnetic and gravitational fields will necessarily have total energy larger than just its rest energy.
Let us now investigate the condition $\mathcal{E}>m$ for a purely radially moving particle. For this particle we can use the four-velocity normalization $g^{\mu\nu}u_\mu u_\nu = -1$ to obtain $$ u_t = -\sqrt{1 - \frac{2M}{r} +\frac{Q^2}{r^2} + \dot{r}^2}\,, $$ where $\dot{r} = u^r = dr/d\tau$ can attain any value. (No speed-of-light violations, you can easily check that $dr/dt \to 1$ as $\dot{r} \to \infty$.)
If we now substitute this into $\mathcal{E}>m$, we easily obtain the necessary condition for escape $$ m\sqrt{1 - \frac{2M}{r} +\frac{Q^2}{r^2} + \dot{r}^2} +\frac{eQ}{r} >m $$ Note that if $e$ and $Q$ are of opposite sign, the $eQ/r$ term is negative and makes it "harder" to make $\mathcal{E}>1$. However, it is obvious that there will always be values of $\dot{r}$ for which $\mathcal{E}>m$. These values are $$ \dot{r}^2 > \frac{2(M - eQ/m)}{r} + \frac{Q^2(1 + e^2/m^2)}{r^2}\,. $$ It is important to remember that this is not a sufficient but just a necessary condition. However, it shows that it is always possible to set the velocity of the particle so that it acquires an "unbound" energy.
The two cases when $\mathcal{E}>m$ but the particle does not reach infinity are the following: First, the case when $\dot{r}<0$, because that corresponds to a radial infall and a geodesic ending in the central singularity never to reach infinity. Second, the case when $\dot{r}>0$ inside the horizon (if it exists), and the particle then escaping outside with $dt/d\tau<0$; this case is in fact the radial infall "played backwards". In any other case $\mathcal{E}>m$ is sufficient for the particle escape.
To conclude, the electromagnetic field does not create an "electromagnetic horizon" which would restrict particles of a certain charge from escaping. The only horizon which puts an ultimate bound on the escape of any particle is the event horizon. As long as we are outside of the event horizon, a particle of any charge can be endowed with enough kinetic energy to escape the potential well.
Here is my attempt to the question:
We have a sphere of charge $Q$ and radius $R$. At a distance $r>R$ the potential $V(r)$ is given by
$$V(r)=\frac Q {4\pi\epsilon_0 r} $$
Now, total energy of the particle X of mass m, charge -q , velocity v is given by $$E=-qV(r)+\gamma m c^2 $$ where $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}$ and observe that when $v\rightarrow c$, $\gamma\rightarrow \infty$
Say escape velocity of X at a distance r is given by $v_e$ and $\gamma(v_e)=\gamma_e$.
From energy conservation if X escapes \begin{align} E(r)&=E(r= \infty)=mc^2 +K.E \geqslant mc^2 \\ \implies -qV(r)+\gamma_e m c^2&= mc^2 \end{align} This gives \begin{align} \mathrm{KE}_\mathrm{initial}&=(\gamma_e-1) m c^2=qV(r)\\ \implies \gamma_e&=\frac {qQ} {4\pi\epsilon_0 r m c^2} + 1\\ \implies v_e &=c\left(1-\frac {1} {\gamma_e^2}\right)^\frac {1} {2} \end{align} This shows if $v\geqslant v_e$ then $\gamma\geqslant \gamma_e$ and X escapes.
But in order to obtain sufficiently high $\gamma$, $v$ just need to be grater than $v_e$ and can also be less than $c$.This shows that if sphere has finite charge Q then for $v_e \leqslant v\leqslant c$, X escapes. Therefore X as well as any particle can escape from any $r>R$ if $v>v_e$ although it is negatively charged without even approaching the speed of light. Thus, the sphere is not a black hole for the negatively charged particles.