Charge, Voltage, or Current? What can be measured with the highest SNR?
In your example, there is absolutely a correct answer, and that answer is current.  Current provides the highest possible signal-to-noise ratio.
Here is why: Johnson–Nyquist noise, commonly referred to simply as thermal noise. All conductors experience this noise, and is the result of thermal agitation of charge carriers within the conductor itself.  In other words, charge is always wiggling around a little bit, and how much they wiggle depends on the temperature of the conductor.
This noise is superimposed on everything else, so it represents a noise floor that cannot be worked around and only lowered by lowering the temperature of anything in the signal path.  Cryocoolers by and large exist for this very reason, but is often not practical or worth the cost.
This thermal noise takes several forms.  The most well known is voltage noise, which is simply roughly white voltage noise with a power spectrum dependent on the temperature and impedance.  Higher impedance or larger resistors will exhibit higher thermal voltage noise.  It is straightforward to calculate the RMS noise:
\$V_{noise}=\sqrt{4k_{B}TR\Delta f}\$
Where \$k_{B}\$ is the Boltzmann constant (\$1.38064852 × 10^{-23} \frac{J}{K}\$), \$T\$ is the temperature in Kelvins, \$R\$ is the resistance in Ω, and \$\Delta f\$ is the bandwidth in Hz over which the noise is measured.
You can also use a handy rule-of-thumb, which is that 50Ω of resistance over 1Hz of bandwidth at room temperature has about 1nV of thermal voltage noise.
Hopefully you've begin to see the problem here.  50Ω is not very much at all, and 1nV is already many many orders of magnitude too high for the purposes of measuring individual ions.  Increasing the voltage we can measure requires increasing the impedance, which just makes the thermal noise that much worse.  Simply put, voltage will not work.
And by extension, neither will charge.
Any solid-state means of measuring charge is going to depend on a known capacitance to do so, but this in itself is a problem.  Recall that a Farad is a Coulomb per volt.  That means that thermal voltage noise will also directly translate into charge noise in any capacitance.  The voltage noise changes the charge in the capacitance, and so you're transferring this same noise into a different measurement.  Since this is an RC circuit, increasing resistance also decreases bandwidth, canceling out the increased thermal noise as far as the capacitor is concerned.  In fact, the resistance term falls out entirely.  What you get is an even simpler equation for charge noise in a capacitor:
\$Q_{noise}=k_{B}TC\$
This is often referred to simply as the kTC noise.  Any capacitance that is practical for metrology purposes and won't be dominated by other mechanical effects changing the capacitance is going to be at best on the order of picofarads.  1 pF has 400 electrons of charge noise.  This is not noise over time, but rather how much the charge will randomly vary from the moment the capacitor is disconnected from an external circuit.
As you can see, charge is far too noisy for our purposes as well.
This leaves current, and with current, we find our salvation.  There is thermal current noise too of course, but importantly, it decreases with both frequency and resistance.  So the ultrahigh input impedance of an electrometer will mean that current noise is extremely low.
As for the circuit, that is honestly beyond the scope of any single question.  What you're really asking is how to build the best electrometer possible.  All Faraday cups fundamentally use an electrometer measuring current.
Typically, these electrometers involve a very low current noise and input bias current op amp, something like the ADA4530-1 or the LMP7721, in a transimpedance amplifier configuration, which amplifies input current into a voltage out, and you can achieve noise figures of roughly 100zA (0.1 fA), as well as sub femtoamp input bias currents, which is sufficient for measurements of hundreds of individual ions or even less.
Ultimately, Faraday cups are limited primarily by the electrometer measuring the current and certain physical effects like secondary emission, which have mitigation methods involving a bias field and other tricks.
Getting that performance is not something you can achieve with a circuit alone, it requires very careful layout and shield rings and general shielding.  All paths for leakage currents must be eliminated.  It is something challenging to someone who has been designing precision analog electronics for years.  Every little aspect of such a design is a question in itself.  Your best bet is to buy a commercial electrometer, or if you really want to build your own, research solid state electrometers.  Just don't expect to match the performance of a real metrology grade piece of equipment easily, quickly, or cheaply.