The electric field of earth is really quite complicated. This paper is old, but presents a nice overview of the issues. That $500\,000\, \mathrm{C}$ number is related a charge separation between the surface and the bottom of the ionosphere, rather than the net charge on the earth. And in any case, it is only the result of a rough model whose underpinnings aren't entirely accurate. Your small sphere is embedded inside this massive system (the Earth's atmosphere). As @John Rennie points out in the comments, because the Earth is so large, you can ignore its curvature, and a better model for this system is a small sphere between the two plates of an enormous capacitor with a $\sim 300\,000\, \mathrm{V}$ potential difference.
So, as always, when dealing with potentials, you need to ask yourself: potential relative to what? Well, assuming that the ionosphere is actually at the same potential as a point infinitely far away, the natural interpretation of your question is to assume that your sphere is neutral infinitely far away, is insulated, transported to the surface of the earth, and then touched to the surface of the earth, and ask how much charge is transferred at this last step. My calculations suggest that a $1\,\mathrm{m}$ sphere has a capacitance of about $100\, \mathrm{pF}$. For a voltage difference of $\sim 300\,000\, \mathrm{V}$, that would be a few times $10^{-5}\,\mathrm{C}$.
Incidentally, as my answer on this question suggests, humans are modeled as capacitors with about the same capacitance, so you would get the same amount of charge.
Of course, these calculations are all for isolated objects, which makes them not perfectly valid, but should be reasonable approximations.
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.
Best Answer
The potential difference
Finding the potential difference is a bit more complicated than you suggested, if I interpreted the meaning of $k$, $q$ and $r$ in your formula correctly.
Before connecting the spheres, they effectively form a capacitor with some capacitance $C$. If $r_1 = r_2$, you can use a formula from wikipedia to calculate $C$, which then gives you the voltage (potential difference) as $$ V = \frac 1C \frac{q_1+q_2}{2}~. $$ If $r_1 \neq r_2$, you will have to solve the Poisson equation, $$ \Delta \phi(\vec r) = - \frac{\rho(\vec r)}{\varepsilon_0}~, $$ where $\phi$ is the electrical potential, $\rho$ is the charge density, $\vec r$ is the position, $\Delta$ is the Laplacian and $\varepsilon_0$ is the dielectric constant of the vacuum. The total charge on each of the spheres serves as boundary condition for the solution. As soon as a solution $\phi(\vec r)$ is found, the potential difference can directly be calculated.
The electric current
Once the voltage is known, the current is, by Ohm's law, simply $$ I = \frac VR~, $$ just as you stated. However this only holds immediately upon connection of the spheres. Afterwards, it makes little sense to assume the charges remain constant, because electric current is just a flow of charge which will eventually cause equal charge density on both spheres and the wire. This is nothing else than the discharge process of a capacitor, which can be modeled quite easily: