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.
A transformer will work in either direction. In each case, the open circuit voltage out the secondary will the that put into the primary times the number of secondary turns divided by the number of primary turns. The only difference is the two windings flip primary and secondary roles when you use the transformer in opposite directions.
You need to look carefully at the impedance the transformer primary presents to the driving circuit, regardles of which direction the transformer is used in. Without load on the secondary, the primary will look just like a inductor. As the secondary is loaded, the primary will look lower impedance. For a ideal transformer, the impedance on the secondary will be reflected on the primary divided by the square of the turns ratio. For example, if the primary has 100 turns and the secondary 200, the turns ratio is 2:1. The open circuit voltage will be multiplied by the turns ratio (it will be 2x of the primary at the secondary), and the impedance tied to the secondary will appear as 1/4 that on the primary.
As long as your source can drive whatever winding you choose as the primary and it does not overload the transformer, it will basically work. Of course there will be losses, but getting into that gets a lot more complicated.
Best Answer
As it's designed for high frequency switching, its primary inductance will be hopelessly low for 50Hz or 60Hz use. So it'll appear as a low impedance shunted across the AC supply or system under test, drawing a high current.
That current, in turn, will saturate the transformer core, further decreasing the inductance - unless the measurement is across a low voltage.
Hopefully, if you are measuring across the AC mains supply, it'll just trip a breaker or blow a fuse before anything worse happens.