This is an edited answer following the remark by Qmechanic pointing out that the capacitance used in the original answer assumes electrical contact between the spheres.
Here I assume that the actual contact time is very small, so no discharge occurs. On, the other hand it is assumed that the motion is slow enough that the laws of electrostatics apply. The capacitance formulas are taken from the article: "Capacitance coefficients of two spheres" by John Lekner. The same notation is used, where the charged sphere is denoted with the subscript $a$ and the other sphere with the subscript $b$. Also CGS units are used (The capacitance has units of length).
Substituting $ Q_a = Q$, $Q_b = 0$, we obtain the following formula for the effective capacitance:
$$ C_s \equiv \frac{V_a}{Q} = \frac{C_{aa}C_{bb}-C_{ab}^2}{C_{bb}},$$
where $s$ is the separation distance. The expressions for the capacitance matrix elements at infinite separations $s=\infty$ are given in the text eq. (8) resulting the following result:
$$ C_{\infty} = R.$$
Of course, this is just the capacitance of the charged sphere.
The capacitance matrix elements become singular when the spheres approach contact $s\to 0$, but using the asymptotic formulas (16)-(18), the effective capacitance has a well defined limit:
$$C_0 = \lim_{u \to 0}\frac{R}{2} \frac{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))^2-(\ln(\frac{2}{u})+\gamma)^2}{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))}= -R (\psi(\frac{1}{2})+\gamma) = 2 R \ln 2,$$
where $\gamma$ is the Euler constant and $\psi$ is the digamma function, and the following identity was used
$$\gamma +\psi(\frac{1}{2}) = -2 \ln 2.$$
By energy conservation
$$2 \frac{m v^2}{2} + \frac{Q^2}{C_0}= \frac{Q^2}{C_{\infty}}.$$
We obtain:
$$v = \sqrt{\frac{Q^2 (2\ln 2-1)}{2m R \ln 2}},$$
which is the speed just after collision in the case of an elastic collision. Of course, the speed is zero in the case (b) of inelastic collision.
