My experience with PFC supplies is mostly theoretical, but as general comments about SMPS, I can say a few things:
Continuous mode sometimes have "prettier" waveforms. This might translate to lower harmonics, if you care. That's why discontinuous mode requires better filtering.
"Correct" selection of modes may be less of a concern. The bigger concern is often staying in one mode, because your controls are all designed around it. Mode transitions tend to be more difficult for controllers to handle.
Boost converters (which share a lot with PFC supplies as I understand them) will enter discontinuous mode at low loads. Which implies that a well-designed continuous-mode controller can probably handle discontinuous mode as well. I'd be surprised if anyone marketed a controller that failed at open load!
If it was me, I'd try the average current continuous mode control (which you seem to be interested in). I'd be surprised if it didn't do what you want. The tradeoffs seem mostly to be about cost in components, which seems to not concern you.
The average value of the output of a buck converter in discontinuous mode is still the same as continuous mode. However, the ripple voltage increases.
This happens on light loads when the switcher cannot produce pulses of low enough duty cycle and it transfers a little too much energy per switching cycle for the light load.
This means the average output voltage starts to rise too high and the control system shuts down the switcher for several cycles resulting in a slightly lower than normal output voltage. Until the load starts to take more power this situation persists.
Discontinuous mode also happens when the input voltage to the buck rises too high.
EDIT - more information:-
If the buck converter is operating at 100kHz and the inductor is 10uH. Let's also say that the supply is 12V and duty cycle is 50%. The "on" pulse of the mosfet will charge the inductor with current and this current is determined by input voltage, output voltage and inductance. Let's say the output voltage is 5V - this means the voltage across the inductor is 7V when the mosfet is "on".
V = L\$\dfrac{di}{dt} \therefore 7 = 10\times 10^{-6} \times\dfrac{di}{dt} \therefore\dfrac{di}{dt} =\space \$700,000
Because dt is 50% of 10us we can calculate I, which is \$700000\times 5\times 10^{-6}\$ = 3.5A.
The energy transferred per cycle is therefore \$\dfrac{L\times 3.5^2}{2}\$ = 57.8uJ.
This transfers 100k times per second \$\therefore\$ the power to the load is \$ 57.8\mu J \times 100000\$ = 5.78W.
If the load resistor is too high to take that power at 5V then the simple fact is that the output voltage will rise and the buck convertor will enter discontinuous mode unless the control loop reduces the duty cycle.
This duty cycle reduction will of course happen because it would be a poor buck converter that couldn't produce less than 50% duty but, it will have a minimum value and at this point, if the energy-load equation isn't in equilibrium the converter will enter discontinuous mode.
Best Answer
A buck converters enters DCM when
$$\frac{2Lf}{R_{load}} < 1-D$$
where
f is the switching frequency
D is the duty cycle
L is the buck inductance
One can replace \$R_{load}\$ by \$\frac{V_{out}}{I_{load}}\$ and one gets the condition
$$\frac{2LfI_{load}}{V_{out}} < 1-D$$
If the duty cycle changes, for example, due to a change in supply voltage, then the converter may enter DCM even though the load current remains constant.
Also, there are some controllers that may alter the switching frequency, with a similar effect.
If \$V_{in}\$ or \$V_{out}\$ or f changes, then the converter will enter DCM at a different current level.
By changing the duty cycle (or frequency) by changing the input voltage.