Let's say we have two copper electrodes as shown in the image, and there is a solution with copper nitrate. In this case, I know that copper will deposit onto the cathode and copper from the anode will be oxidized and enter solution when a voltage is applied across the two electrodes. However, what if instead of copper nitrate, there was just NaCl (or any other electrolyte that wouldn't cause any problems like precipitation) in solution instead? Would current still be able to flow, and if so, would the electrodes still lose and gain mass as they did before? I am fairly certain that current does flow, but I am not really sure if copper will still deposit onto the cathode and vice versa.
It is a complex problem. Yes, NaCl solution is a conductor, so electricity will pass through the solution. Now in your situation both electrode are made of copper. Since electricity is passing through the solution, electrolysis must occur at both electrodes. You cannot have electrolysis only at one electrode.
Assuming there is sufficient salt in the solution, what will happen at the anode?
Chlorine gas will evolve. I will let you find relevant equations. Chlorine gas attacks copper very quickly and the anode starts to "degrade" quite fast, but will it pass into solution? Nope.
Let us see what is happening at the cathode?
Sodium ions cannot be reduced in water esp. on copper electrodes (on mercury they do), instead water will be reduced. What happens when water is reduced? It turns into hydrogen gas, which escapes, and leaves hydroxide ions behind. This process makes the solution basic.
Now coming back to the anode, since you have hydroxyl ions around, they precipitate copper ions. All you get is yellow orange junk in such a system. Cu(II) is blue so why an orange yellow mess? It is most likely due to Cu(I) compounds. Anode dissolves pretty fast but nothing happens to the cathode.