In theory, yes, because there is no virtual ground at the summing point (the main benefit of the inverting summing stage), and the signals would go back through the other channels summing resistors and interact with where the pot is positioned, as far as sinking current into either the low impedance Vref source or the low impedance channel buffer opamp output, or having some added resistance from the pot in the way because the wiper is somewhere in the middle.
In practice, no, short answer is because the summing resistors are big and the other channels pots moving around are a small change in load.
Long answer is because the summing resistors are much larger than the pots, worst case, a pot from Channel B is right in the middle of its resistance, and the Channel A signal sees it through the 100K resistor from its channel to the summing point, then another 100K resistor back to Channel B's pot, as a 5K resistor to the opamp output (Lo-Z, with ground-like current sinkability) and a 5K resistor to Vref (Lo-z, ground-like), so those combined, behave at most like a 2.5K load in the middle of the pots resistance and less as you move to either side, which with the 100K summing resistor from channel A to the summing point, then through the 100K summing resistor to Channel B's pot wiper, is a small variation in attenuating load on Channel A's signal at the summing point, 102.5K vs 100K. Doing the voltage divider math, at most 1.395% variation is possible in the volume of one channel if you wiggle the all the other channel's volume pots around. Since to the human ear something has to be twice as loud to be perceived as a noticeable difference, these tiny variations are not an issue.