There will be some aluminum in the vapor phase and it will condense with the water: it will be a very tiny amount. It does not actually matter what the container is, there will be some vapor emitted. Just a question of how much. Most solids have a very low volatility, meaning that there is practically nothing that is lost to vapor from the surface of the solid. This is why chemical experiments or measurements tend to use doubly-distilled water at minimum as the dilutant or solvent: there is less dissolved solid in the second pass than the first. The first distillate still has measurable dissolved salts, often down at the very low ppm to ppb level only.
The amount of metal that you will get from the condensate of water vapor coming from an aluminum container (pot or kettle) will be tiny and hard to measure without very good analytical equipment. there will be some aluminum there, but it will be a very tiny amount.
It is almost impossible to make a truly pure substance. However, when you are dealing with something on the order of 10^23 atoms for every 20-100 g of material, "Nothing" (we can't measure it level) is possibly as much as 10^12 atoms or thereabouts for that mass. Sounds like a lot but it isn't at all, that would be fractional parts per trillion in concentration - typical measurement methods won't see it. We all have some PCB and dioxin in us, but the concentrations of those nasty chemicals is way down there (parts per quadrillion). Still there though. The stuff is everywhere.
So, yes, there will be aluminum (impossible to avoid) but no it won't be enough to matter. It is possible to vaporize some of the pan from the heating (we can smell a hot pan and what do you think is the source of the metallic smell?) and even transfer enough so that you could detect it in the primary distillate, but even then it would be a tiny amount, not a major deal.