Your microwave creates a standing electromagnetic wave inside itself but it doesn't consume much energy to create this wave, so if the oven is empty it will only consume a small amount of power - a perfect microwave oven would consume no power when empty. Power is only consumed when you put something in the oven that absorbs energy from the standing wave. The oven needs to take energy from the mains to replace the energy absorbed by whatever's inside it.
The rate at which something inside the oven absorbs energy depends on what you put in. For example if you put in a cup of cooking oil it will absorb energy slowly and heat up slowly (I wouldn't try this at home as traces of water in the oil can boil explosively!!).
The 750W rating of the oven doesn't mean it pours 750W into whatever is inside it; it means that's the maximum amount of power it can pour in. The actual power absorbed will typically be less than this and possibly much less.
It would be an interesting experiment to try heating two (or more) mugs of coffee at the same time. I would bet you'll find the total energy absorbed increases as you put more mugs in, up to the 750W limit.
Later:
Prompted by Anna I have done the experiment. I used two identical coffee mugs containing 400g of water each. I first heated just one mug and measured the temperature rise every ten seconds from about 10C up to about 25C - I didn't want to go higher because you have to start worrying about heat loss to the mug and air.
I then replaced the water and heated the two mugs together, spaced as widely apart as possible, and again graphed the temperature as a function of time again up to about 25C. The results? Well my oven is rated at 600W and with one mug I measured the rate of temperature rise to be 481W (plus or minus a few percent). With two mugs the rate of temperature rise was 530W.
Now I'll just post the results up to the arxiv :-)
I would side with your colleague.
When you mix, the energy is constant so the following is fulfilled (first law of thermodynamics) $$\Delta Q_{water} = \Delta Q_{cup} + \Delta Q_{milk}$$
So, there is a flow of energy (heat $\Delta Q$) from the water to the cup and milk. This flux will stop when all temperatures are the same.
In this picture (closed system of water, milk and cup and long time), it wouldn't matter how you mix and you would get always the same temperature.
Now, a caveat. Here there are no losses and in practical terms the transfer of energy is instantaneous (which would be equivalent as waiting "infinite" time). My bet is that such effects are very small and shouldn't change the conclusion unless you have a very strange cup (material-wise) or very little water and milk.
Best Answer
It's probably because your milk cup is made of a material that is a relatively good thermal insulator.
First of all, the microwaves directly heat the milk, and not the cup, as long as the cup is made of material that microwaves pass through without being absorbed.
The heated milk, in turn, being in contact with the sides of the milk cup directly heats the sides but does not directly heat the handle because the milk is not in direct contact with the grasped portion of the handle. For the handle to get warm there needs to be heat transfer by conduction from the sides of the cup to the handle.
The heat transfer from the sides of the cup to the grasped portion of the handle will depend on the thermal conductivity of the cup material, the cross sectional area of the handle part of the cup, and the length of the path from the side of the coffee cup to the part of the handle being grasped. If your milk cup is made of glass, it's thermal conductivity is relatively low (roughly 1 W/m K) making it a reasonable thermal insulator.
All of the above factors can keep the grasped portion of the handle from getting too hot regardless of how hot the milk is, at least for a reasonable amount of time.
Hope this helps.