There isn't really a "temperature five feet off the surface" because there isn't much of anything there to have a temperature. The Moon does have an "atmosphere" but typical pressure is around $0.3 nPa$ (wikipedia) and it will not tranfer enough heat to or from any reasonable thermometer to produce anything you could call an ambient temperature. Indeed the mean free path of molecules in such a gas is measured in kilometers, so they will interact with the surface more than with each other, making the whole question of temperature a bit moot.
What your thermometer will measure is the balance of radiant energy that it is absorbing with what it is emitting. That is, the temperature it reports will be the one at which there are in balance. The depends on exactly what is in line-of-sight of the thermometer, and also on how well it absorbs and emits different frequencies. If it has no line-of-sight on sunlit (or recently sunlit) rocks, or on the Earth it could get very cold indeed, since it is absorbing very little and can radiate into space. On the other hand, if it almost surrounded by warm rocks that either are, or recently have been heated by the sun, it will end up almost as hot as they are.
Based on the more exact specification added to the question: Assuming you are on a flattish bit of Moon, then basically you have to ask where the lines of sight from the glass of water go. Nearly half of them (going down) will hit the lunar surface. Most of the rest will go into empty space. A few may hit the Earth, and a number will hit the curtain. It's a little simpler if we replace the curtain by a mirror, reflecting the Sun's heat away rather than absorbing it. To a VERY VERY rough approximation then, half of what we hit is surface at 400K and half is space at 2.7K. Energy transfer due to radiation scales as the difference of the fourth power of temperature, so we get:
$$400^4-T^4 = T^4-2.7^4$$ (wikipedia)
with a solution around 336. So the water would not freeze, although it wouldn't boil either.
Among many other details, this solutions assumes that the water and the rocks are black bodies, absorbing and emitting evenly and effectively at all wavelengths and that the glass is transparent at all relevant wavelengths.
Edited to use correct equation, and consequently change the outcome. Thanks to commenters who pointed out my mistake.