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.
It should be associated with the work function, which is the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface, and different materials have different work functions.
Consider a very simple case, that a spherical electrical object exists in vacuum.
Considering that we move an electron from the object to the infinitely distance, if the energy of the system decrease, then it shows the electrical object is unstable. Hence, the maximum electric charge that an object can hold should make the energy unchanged in the process of removing electron.
If its radius is $R$, the work function is $W$, the maximum electric charge that it can hold is $Q$, then $$\frac{1}{4 \pi \epsilon_0} \frac{Qe}{R}=W$$
the left side is the change of electrical energy.
As for the object with other shape, you have to change the form of expression in the left side.
Best Answer
Avogadro's number is $6.02\cdot 10^{23}$; a single electron has a charge of $1.6\cdot 10^{-19}$ C, so $1.04\cdot 10^{-5}$ moles of single-ionized material carries a net charge of 1 C.
To carry that much charge, you need a large capacitance or a large voltage, since $Q=CV$. An object with a 1 mF capacitance and a voltage of 1000 V would be sufficient, or a 1 F capacitance with 1 V. The difference between these is the amount of energy stored, which goes as $E=\frac12 CV^2 = \frac{Q^2}{2C}$ - so for the same amount of charge, a larger capacitor will have less stored energy.
And that is the hint to the "will it kill me" part of the question: you are not killed by charge, but by current flowing. If you have a charged object with a low potential, the flow of current through your body will be slow - and you will survive. But if the voltage is high, it will easily overcome the resistance of your skin and give you an almighty jolt - possibly enough to kill you.
So what is the size of a sphere with a capacitance of 1 F? Capacitance of a sphere is $4\pi\epsilon_0 r$, so you would need a radius of about $9\cdot 10^9 $ m - quite a bit bigger than the Earth. A sphere with a 1 m radius, with a 1 C charge on it, would have a potential of about 1 GV. That's a very large voltage - if you could even maintain that potential (not in ordinary atmosphere), touching it would kill you.
Supercapacitors can be created in which two conductors are brought in very close proximity, while having a dielectric layer in between that produces a very high capacitance in a small package. Such a device can easily be charged with a Coulomb - although that isn't a net charge (one plate will be positive, the other negative). And whether such a capacitor could give you a lethal shock will again depend on the capacitance.