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.
The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is:
$$ F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2} $$
and the field strength at the surface of the second sphere is:
$$ F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2} $$
Lets take the ratio $F_1/F_2$ to see which is greater. The constants cancel to give us:
$$ \frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}} $$
and I'm going to rewrite this slightly to make it obvious how you use your equality $Q_1/r_1 = Q_2/r_2$:
$$ \frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}} $$
Because $Q_1/r_1 = Q_2/r_2$ we can cancel them on the top and bottom of the fraction and we're left with:
$$ \frac{F_1}{F_2} = \frac{r_2}{r_1} $$
and because $r_2 < r_1$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1.
and 
to come up with 
but I believe this is for a single sphere, and am unsure how the second sphere would affect my results. Do I ignore the second sphere altogether?
Best Answer
You have an expression for the total energy
$$U_E = \frac{1}{2}\int_V \rho(\vec{r})\phi(\vec{r})\mathrm{d}^3x.$$
And now you can break the charge density into two parts $\rho=\rho_1+\rho_2$ and the potential into two parts $\phi=\phi_1+\phi_2$, where each is due to just one sphere (so the additions hold by linearity of charge and superposition of fields/potential). Put them in and you get
$$U_E = \frac{1}{2}\int_V (\rho_1(\vec{r})+\rho_2(\vec{r}))(\phi_1(\vec{r})+\phi_2(\vec{r}))\mathrm{d}^3x.$$
Which equals
$$U_E = \frac{1}{2}\int_V \rho_1(\vec{r})\phi_1(\vec{r})+\rho_2(\vec{r})\phi_2(\vec{r})+ \rho_2(\vec{r})\phi_1(\vec{r})+ \rho_1(\vec{r})\phi_2(\vec{r})\mathrm{d}^3x.$$
The first two terms integrate out to be the energy of the spheres in isolation. So the last two terms are the interaction energy:
$$U_{int} = \frac{1}{2}\int_V \rho_2(\vec{r})\phi_1(\vec{r})+ \rho_1(\vec{r})\phi_2(\vec{r})\mathrm{d}^3x.$$
Now, each term integrates to tell you the total work done by one sphere on the other sphere as you (hypothetically) bring the isolated objects from super far away to their current positions. Since the forces were equal and opposite, the work done by one on the other is equal to the work done by the other on the one.
So the remaining two terms actually give equal integrals.
$$U_{int} = \frac{1}{2}\int_V \rho_2(\vec{r})\phi_1(\vec{r})+ \rho_1(\vec{r})\phi_2(\vec{r})\mathrm{d}^3x= \int_V \rho_1(\vec{r})\phi_2(\vec{r})\mathrm{d}^3x= \int_V \rho_2(\vec{r})\phi_1(\vec{r})\mathrm{d}^3x.$$