In my experience, no, and as I understand the theory, no. So, I THINK that should just mean, no. I'm sure PRR will be along to explain why I'm wrong at some point, though.
Theory is almost always based on experience, or modified/verified by experience. So I doubt PRR will say you're wrong.
Look at the attached diagram. It is a single pickup, with a tone circuit. I've ignored the pickup switch, showing just the complete circuit for the selected pickup.
Where's the ground? Where's hot? I've omitted any ground symbol, and connected the points that might have a ground symbol with a "wire" which is exactly what happens in the physical circuit.
Ground is just a reference point, not a magic black hole where noise disappears. You could choose to use a different point in the circuit (derived from the word "circle") as your reference. Would it make any difference in the possible noise where you placed your reference?
The way shielding works on an unbalanced circuit is that the shield, which is part of the signal-carrying circuit, is connected to your reference. The hope is that any noise induced in the hot conductor also gets induced in the reference, so both are altered the same way. Since there is (hopefully) no difference between the two except for the desired signal, there is no noise.
It's like you and a friend standing on a boat. If the waves cause the boat to bob up and down, you might notice when you look at the water or the shore. But if your only reference was the boat and your friend, you don't notice. You're both bobbling at the same time by the same amount.