In reality this circuit will weld the reed to the left contact the first time it charges if the capacitor is of any sensible size for this task. In theory, charges and discharges "fully" take forever. But it will mostly charge in a multiple of R*C that is large enough, say 5 or 10. So it still makes sense to try to solve this.
One way (maybe the best way) to handle this problem is to consider energy. Each time the capacitor is charged from 0V to the battery voltage it gains energy
E = \$\frac {CV_B^2}{2}\$. Each time it discharges "fully" it loses that energy.
If the cycle repeats at a periodic frequency f we know that the power transfer is f*E. So the RMS current through the ammeter must be:
\$I_{RMS} = V_B\sqrt{\frac{fC}{2R}} \$
Since capacitors in parallel add, doubling the capacitance will increase the RMS current by \$\sqrt{2}\$, all other things being equal. You can show that by using the energy equation for each capacitor.
On the other hand, the average current will double.. and it has a particularly simple equation Iavg = f * C * Vb. (R drops out). Proof left to the student.
Note that the current in the resistor will follow an exponential discharge curve from Vb towards zero with time constant RC each cycle, then drop from a very small number to exactly zero when the switch opens.
The impedance (think of it as resistance) of a capacitor changes with the frequency of the signal passing through. The lower the frequency (bass sounds) the higher the impedance.
The impedance of the capacitor also depends on its value. A capacitor with a higher value will have a lower impedance than a capacitor with a lower value. For the same frequency, a small valued capacitor represents more resistance than the large value capacitor.
In order to get more bass, you have to use a larger capacitor in series with the speaker.
C1 in your circuit is there to block DC from the amplifier. At DC, a capacitor is very close to an open circuit - DC cannot pass.
The change over is gradual, however. The capacitor doesn't just block DC. It also impedes the flow of other frequencies. The lower the frequency, the more it is blocked.
At some point it is no longer noticeable. For working with filters (the capacitor/speaker combination is a high pass filter,) this point is defined as the point where the amplitude is reduced by half (that's -3dB.)
I'm not going to get into calculating the cutoff of a filter - there's plenty of explanations on the web that go into much more detail than I want to.
For the other side (resistor changes sound,) we have to look at inductors.
The pickups on your guitar are inductors - basically just coils of wire.
Inductors are the opposite of capacitors. Inductors let DC pass just fine, but their impedance goes up the higher the frequency. It also goes up as the value of the inductor increases.
You aren't changing the impedance of the inductor (pickup.)
When you change the resistor at the amplifier, you are changing the load on the inductor.
A resistor that is connected across the inductor forms a voltage divider. How the voltage is divided between the pickup and the resistor depends on the frequency of the signal - the impedance of the inductor changes with frequency which changes how the voltage is split between the inductor and resistor.
The combination of the coil and the resistor forms a low pass filter. It removes high frequencies.
The point (frequency) where this begins to be noticeable depends on the resistor loading the coil. A higher value resistor allows more high frequencies to pass. Lowering the value of the resistor lowers the frequency at which you can hear a difference.
Another thing that will happen is that the resistor also changes the amplitude of the signal presented to the amplifier. A higher resistor means less signal getting to the amplifier, which results in a quieter output.
A lower resistance means more signal to the amplifier, which gives a louder output.
For a guitar player, there is also the interesting possibility of distortion. You provide so much of an input signal that producing the amplified signal would require more voltage than the power supply of the amplifier.
When that happens, the output voltage will "stick" to the powersupply voltage until the input signal is smaller.
This is known as clipping, and is a bad thing in a general amplifier, but can be a useful thing for a guitar player.
Best Answer
They both would work if they are within the voltage range. However, they do not work the same. Of course, besides dimension differences, there are different parameters that change when having different voltages and also depending on the capacitor type.
For ceramic capacitors you have to consider the bias voltage (VDC). That is, when increasing the voltage the capacitance decreases. You can see the difference between two 100 µF capacitors, one rated at 25 V and the other to 6.3 V. Both would support 5 V, but the capacitance of the 6.3 V capacitor would be lowered (around 50%). The bias voltage is actually dependent on size rather than the capacitor voltage level, but it is true that higher voltage levels usually means higher volume.
Aluminium electrolytic capacitors are not as vulnerable to the bias voltage problem as ceramic ones. But capacitors with a lower voltage rating usually have higher ESR. That is something that might be considered (especially if there are no ceramic capacitors that have a very low ESR around). Also, tolerance values are usually worse for a higher voltage level.
So in general they would work the same for most applications. But there are some issues to consider when choosing the voltage level. The safest choice would most likely be to have just the voltage level needed for your application with some margin. But going with higher voltage levels, like in your case, is not usually a problem. It is more critical when going for lower voltage levels. Here you have a small recap of different capacitor types.
Sources:
Graph: Murata Electronics.
Table: NIC Components.