My first reaction is that trying to externally turn off the micro is the wrong way to go about this. Perhaps you are using the wrong micro, but there are micros that take very little power when in full sleep mode. Take a look at some of the "nanoWatt" (marketing term) PICs and MSP430s. The latest PICs are basically down to a tiny amount of leakage current in sleep, less than 1uA for some.
How low a current do you need? What does the rest of the circuit draw. What is the CPLD current when it's not switching? It's hard to give a good answer without some real numbers.
Saying that something is a "huge issue" is no spec at all. For example, if you are trying to run something as long as possbile on a CR2032 battery, then 1uA sleep current is fine since the effective self-discharge current is more than that. If you want 3 years from a single AA battery, then even more would be acceptable.
EDIT: Something else I should have added. If you really are going to switch power to the micro (I still think that's a bad idea, get the right micro instead), you probably need to switch the ground instead of the power. You say there are IIC and UART lines connected to the micro. IIC has passive pullups, so these will either draw current or power up the micro thru the protection diodes if you try to switch off the power instead of the ground. Logic level UART signals idle high, so there could be a similar issue there too.
In any case, you can apparently redesign the board, so I don't understand how you're stuck on that particular micro, whatever it is. If power is really such a "huge issue", then everything else should be on the table.
The 20v voltage which you see across 2 chassis is high impedance sourced from mains. The schematics is a diamond shaped bridge fromed by 4 capacitors, real and parasitic. Top and bottom of bridge is mains. Horisontal of bridge is your voltmeter. Shoulders of bridge is this small ceramic UL certified capacitors, capacitive coupling of power cable to chassis, intercoil capacitance of power transformers. There is also an inductive background. The non-zero figures on voltmeter are caused by asymmetry.
If your equipment (power supplies) is generally trusty, you can tie grounds safely, so the few microamperes AC will be shorted, and it will resolve the problem.
But in your case you use something home brewed in combination with insulated device: modem is insulated by design with output transformer. So, no easy solution.
First class solution is creating lab grade ground using real earth and adding isolating transformer. Check the local electrical code as well. Only after this you can tie grounds.
Best Answer
Your question is not totally clear, but I assume that your uc must measure the 48V, not just know whether it is (within some wide margin) present at all.
First: Is it a problem if the 48V and the uc share the ground? If so, you will need isolation and you will have an interesting design challenge.
If a common ground is no problem, I see no need for isolation (unless there are considerations that you did not tell us about).
You need an input voltage range that your uc can handle, with an impedance that is not too high for the A/D input. You can use a resistor divider: choose the lower resistor equal to the highest impedance that your A/D allows; choose the upper resistor to divide the maximum of your range-of-interest (for instance 30 .. 60V) down to what your A/D input can measure.
If the 48V can be higher than the highest value you considered you can either take that value into account when calculating the upper resistor (which reduces your accuracy), or add some overvoltage protection on your A/D input (like a zener clamp, or skottky diode clamp to Vcc if it can handle the extra current).
If you or your safety guys are worried about the consequences of a failure of the upper resistor you can split it into two equal values, and design the clamp to handle the current when one of the two shorts.
Do note that the accuracy of your measurement will be lower for lower '48'V voltages, and it will depend on the accuracy of your reference voltage (often the Vcc of the uc).