If your load is powered by a regulated switching power supply, then what you were told could be correct.
Internally, this device is trying to produce, for example, 5 V to power the actual circuits inside. And those circuits might need (as an example) 10 A. So the final load needs 50 W of power.
The switching power supply circuit is designed to adjust the current it draws from its input to whatever it needs to be to provide 5 V to its output.
So if you supply it 120 V, it will draw about 420 mA. If you use a too-small cable and only 100 V is available at the input to the power supply, then it will increase its current draw to 500 mA to be able to deliver 50 W to its load. (I'm ignoring the inefficiency that you'd find in any real power supply circuit that would make the required input power slightly higher than the output power)
Obviously I've used smaller numbers in my example than you are dealing with in your actual situation, since 500 mA is not going to throw a lot of breakers. But the general principle is the same. A higher voltage drop in the cabling to your device can lead to the device drawing more current.
If the load is a motor, again your advisor's comment could be correct, but for different reasons. If the supply wires drop too much voltage, then the motor will be more likely to stall, and when it stalls it will dramatically increase its current draw (which is likely to throw a breaker, as in your advisor's example). Lowering the voltage available to the motor could also increase the current draw if it's not stalled but operating with the same rpm and mechanical load (but I suspect whether it would maintain the same rpm depends on the kind of motor and the kind of controller used with it, about which I don't know enough to talk in detail).