If "materializing" means creating matter from energy, then a 250 ml cup of water will contain approximately 250 g of water with a smidge of other molecules. Add another 100 g for the cup (light weight... this is space). From $E=mc^2$, you find an energy content of roughly $3\cdot 10^{15}~\rm{ J}$.
Note that the thermal energy content (difference between "cold" and "hot") is many orders of magnitude smaller than this - completely negligible.
If we assume the mass of the Enterprise to be about $3\cdot10^9~\rm{kg}$ (source), this corresponds to the kinetic energy of the entire starship moving at 3000 m/s (Mach 10).
I'm thinking that it would be cheaper to get a Keurig...
The fact that sugar does not change the temperature can be true but this means that the time it takes to dissolve the sugar is important. The coffee will cool down exponentially all the time. Then, you drop the milk, which additionally reduces the temperature. But, mixing milk with coffee takes no time. So it will me immediate drop of temperature by some degree.
You can also drop both together and wait until sugar dissolves in the colder beverage. I think this is exactly the flipped order scenario. The milk, dropped into the hotter tea, causes more degrees to drop immediately because, say, when it is as cold as milk, the milk won't cool it down at all. I think that this is because you have an average energy between two masses, of tea and milk. Temperature is proportional to the energy, $$T_\textrm{mixed} = {MT_M + mT_m\over M+m}= {kmT_M+mT_m \over km+m}= {kT_M + T_m\over k+1}.$$
I think that sugar-first gives you higher temperature because in the first scenario you have shorter period of sugar melting, followed by some degrees lost due to milk. In second scenario, you first lose more degrees due to milk and then will have longer period of sugar melting, which again makes more degrees lost during the melting stage (compared to melting state in the scenario 1). So, since every of two phases causes more dramatic temperature loss in scenario №2, you will end up in colder liquid.
It is difficult to prove formally.
In the first scenario, you have t1 seconds to dissolve the sugar so that temperature drops from T10 to T11 and then additionally drops by T1m.
In the second, initial T20 is reduced by milk to T20-T2m which further exponentially reaches some end temperature during t2 seconds.
The temperature is decaying exponentially, proportionally to the difference between cup and room temperature, something like $T(T_0, t) = T_r + (T_0-T_r) e^{-at}$ where $T_r$ is the room temperature and t is current time since the beginning of experiment.
Let's have some amount of sugar m, which is solving at speed dm/dt = b*T(t), that is, we loose the sugar proportionally to the temperature. Integrating over time, we can figure out when all sugar is dissolved, $$\int_0^t kT(T_\textrm{fresh}, t)dt = \int_0^t T_r dt + \int_0^t T_r dt + \Delta T \int_0^t e^{-at}dt = T_rt + \Delta T {b\over a}(1-e^{-at}) =m.$$ We need to find out the $t$ such that the dissolved amount of sugar amounts to $m$. For me, $at + e^{bt} = c$ is quite difficult to solve for $t$. But, if you can, we know the time it takes to dissolve the sugar and can proceed and find the temperature we have at that point, $T_\textrm{sweet} = T(T_\textrm{fresh}, t)$ and finally $T_\text{sugar+milk}=(kT_\textrm{sweet}+T_m)/(k+1)$.
In the second scenario, we first find $T_\textrm{milked} = (kT_\textrm{fresh}+T_m)/(k+1)$, compute sweetening time $t_2$ given $T_r t_2 + (T_\textrm{milked}-T_r){b\over a}(1-e^{at_2}) =m $, taking temperature after $t_2$ seconds, $T_\text{milk + sugar} = T(T_\textrm{milked}, t_2)$
I can hardly contrast $T_\text{sugar+milk}$ with $T_\text{milk + sugar}$.
Best Answer
Dissolving depends on several parameters, one of them is the amount of matter already dissolved in the medium. You can't dissolve any amount of sugar in a cup of tea, at some point it stops dissolving.
When circling, the stuff moves all the time pretty the same path thus you don't mix sugar and water too much comparing to each other, they move the same direction because of inertia. Thus the concentration of sugar in some parts of water may be higher and sugar dissolves slower.
In back-and-forth "style", you always change the direction the particles move and thus you mix them more effectively, so you can't have areas where sugar is more concentrated. Which means it should dissolve faster.
That's my assumption :)