In answer to the title question, "yes, but...." (it's not practical and the effect is too small to be noticed in the sort of situation you describe)
1 - whatever energy you use to stir the water, ends up as heat pretty soon (which raises the temperature)
2 - it is a good idea to develop an intuitive feel for the magnitude of mechanical versus heat energy. There is a lot of mechanical work in a "small" (in the everyday sense) amount of heat. For example let's say your bucket has 10 litres (10kg) of water, and its temperature falls by 20 degrees-C in 10 minutes (for the purpose of illustration). That's an average temp drop of 2degrees-C per minute, which means that via the various heat loss mechanisms (conduction, convective transfer to air, vapourization and transport, etc), heat is leaving the water in the bucket at rate of 83680 Joules per 60 seconds (2 degrees C times a heat capacity of 4184 Joules/deg-C/kg times 10kg of water), which is 1395 Joules per second which is 1395 Watts. So for you to keep the temperature of the water constant you would have to add mechanical work to the water at this rate, which is an absolutely huge amount for a person to deliver. This is about 1.8 horsepower; perhaps a topnotch cyclist, delivering sprint-level output on an exercise bike, might be able to produce this power level for a short period of time. The important point is that this is several orders of magnitude more than what vigourous stirring by hand might deliver.
For what it's worth heat from pumping is real, for example my neighbour has a small hot tub and its heating comes solely from the circulation pump motor. I haven't been able to find a photo on the internet of that sort of setup but here is mention ("pump friction") of the method:
http://www.precisionspa.com/Precision-Spa/Spa-Heater-Repair/spa-repair-hot-tub-heating.html
I think your reasoning is sound.
However, being in Melbourne, Australia (which is heading for 35 C today), I can think of better applications of eating ice than burning calories. You would certainly experience a greater drop in body temperature by consuming equal amounts of ice versus water at 0 C.
Best Answer
The motion of molecules that is responsible for heat content in water is random motion; that is there are molecules moving in all directions. The directed motion that you are considering ( all molecules moving in the same direction) from the flowing water does possess kinetic energy, but it is not heat energy.
However, if the water flow encountered some obstacle that impeded the flow and randomized the motion of the molecules, some heating would occur. In that case some of the kinetic energy of the flow would be converted to heat energy ( internal energy) of the water.
It may be worth noting though that the average magnitude of the random velocities of the individual water molecules is quite high (probably on the order of 500-600 m/sec for room temp water ) compared to the likely velocity of any "fast moving water current", so the heat energy or internal energy would be large compared to the kinetic energy of the water current as well. Thus, not much heating would be expected from converting the flow to randomized motion.