No problem, glad I could provide some helpful information, to you and anyone who comes after. No doubt we’ll end up refining it a bit and turning it into a guide on the main website so it’s easier for others to find later 
Thanks for the share - good stuff. That goes quite nicely into the maths, which complements the more intuition-focused explanation I provided
You can use static analysis to determine the location of an equivalent point mass applied at the CoM, as below:
The object isn’t moving, so you know that all forces and moments/torques equate to zero. Taking the vertical forces due to gravity, with corresponding reaction forces,
The same applies to the moment (rotational reaction), so considering the moment about point A,
Note that scales return mass rather than force, but if everything gets divided by a constant (earth’s gravity in this case) the relationships still hold.
As a couple of examples, consider the following cases:
For the uniform mass distribution you just end up with x_{CoM} = x/2 (which matches intuition), then in the uneven distribution case you get x_{CoM} = x \cdot \frac{50}{130}, so the CoM is 38.46% of the way across from the left.
One measurement with two scales constrains the CoM to a plane, so you can then rotate 90 degrees about each axis to get the other two planes, and the CoM is the point intersection of the three planes you determine. Alternatively you can do two measurements with four scales, and just pair up the scales for the calculations, and you should end up with redundant information for one axis (e.g. if you have scales S_1, S_2, S_3, S_4 in a rectangle you can take a measurement with the ROV sitting flat, then do R_1 = S_1 + S_2,\ R_2 = S_3 + S_4 for one axis, then R_1 = S_1 + S_4,\ R_2 = S_2 + S_3 for the second axis, then rotate the ROV from flat to lying on its side for the second measurement, and then you only need to do the missing axis (since one of them was done by the first measurement), or you can do the calculation anyway as a consistency check).
To anyone curious, CoG is the same as CoM when in a constant gravitational field, which we can definitely assume is the case of an ROV operating on earth. Significant differences only occur when the object of study is similar in scale to the source of the gravitational field, or if the object of study is between one or more gravitational sources with a similar influence at the object’s location.
To answer the question, water exerts pressure but that’s unrelated to CoG. The gravitational force can change somewhat if you get close enough to earth’s centre that there’s enough mass above that it, but given earth’s radius of ~6370km that’s not feasible in an ROV, particularly given the ocean’s deepest point is all of ~11km down. On top of that, ROVs are far too small and tightly bundled for the CoG to change much with respect to the CoM anyway, so yes, for practical purposes CoM and CoG are equivalent here