Metastable water vapor exists as supercooled vapor or in supersaturated air e.g. in the absence of cloud condensation nuclei.
At room temperature and a pressure of 1 atm supersaturated air exists, but I doubt that it is possible to prepare pure supercooled water vapor at this conditions.
The usual phase diagram of water describes water as a pure substance, not as an n-ary mixture with gases. According to this phase diagram, water is liquid at $20\mathrm{°C}$ and $1~\mathrm{atm}$ ($1013.25~\mathrm{hPa}$).
However, when you follow the line $T = 20\mathrm{°C}$, you will find an intersection with the curve that separates the vapor and liquid phase. The pressure at this point ($23~\mathrm{hPa}$) is called the vapor pressure at $20\mathrm{°C}$. This is approximately the same as the partial pressure of water in saturated air at the same temperature. So you can think of air molecules as mere spectators that do not contribute to the "water pressure".
Conclusion:
The answer would be no under the additional assumption that the air is saturated with water vapor. However if air is supersatured with water vapor then the water vapor would be in a mestastable state and the answer would be yes.
Where does an evaporating molecule get enough energy to escape?
In any system of a large number of molecules, the molecules are going to have a variety of kinetic energies (even if they all start out at the same kinetic energy, random collisions transfer energy and ensure that a few molecules "get lucky" by being boosted in the same direction multiple times). Based purely on the statistics of such systems, we have a relatively good idea of the distribution of kinetic energy among the particles of a fluid. This distribution is called the Maxwell-Boltzmann distribution*. Two examples of this are shown below (the vertical axis can be thought of as the relative number of particles having a particular speed/kinetic energy):
As you can see, there is a "tail" of this distribution that indicates that a few particles have very high kinetic energy, and these are the particles that escape the liquid. You can also see that at higher temperatures, the fraction of particles with very high kinetic energies substantially increases, which is why evaporation happens more quickly at higher temperatures.
*The Maxwell-Boltzmann distribution is technically only valid for an ideal gas; liquids have additional interactions (like van der Waals forces and, in the case of water, hydrogen bonding) that add correction terms to the Maxwell-Boltzmann distribution. In water, for example, substantial hydrogen bonding reduces the extent of the high-energy tail and causes water to evaporate slower than expected from this picture alone. Still, though, qualitatively the Maxwell-Boltzmann distribution provides the correct intuition.
How is this process affected by humidity?
In a container containing a liquid, there are actually two opposing processes at work: high-energy molecules of the liquid escape the surface and evaporate, and low-energy particles of vapor condense into the liquid. Humidity is the density of vapor, so evaporation works to increase the humidity, and condensation works to decrease it. There is always an equilibrium level of humidity where evaporation and condensation occur at the same rate, and the amount of liquid and vapor do not change with time. If the humidity is lower than this level, then evaporation will occur faster than condensation, and the net effect will be that the liquid evaporates; if the humidity is higher than the equilibrium level, then evaporation will occur slower than condensation, and the net effect will be that some of the vapor condenses. The equilibrium level of humidity is affected by temperature, because it depends on the ratio of high-energy molecules in the liquid to low-energy molecules in the vapor. High humidity doesn't prevent high-energy molecules in the liquid from escaping; it just ensures that more molecules are entering the liquid than are escaping it.
Wouldn't the air molecules bump into the water molecules and cause them to go downward, not upward?
Sometimes they do. But the density of the vapor above the liquid is always lower than the density of the liquid itself, so it's not particularly likely that a vapor molecule is in the right position to do this, and high-energy molecules leaving the liquid usually end up entering the gas.
Best Answer
Yes. If the pressure is low enough, the net rate at which molecules in the liquid will leave the liquid phase will be so great the liquid will boil.
Above that pressure, the rapidity with which gaseous molecules of the liquid species are moved away from near the surface will affect the rate of evaporation.
Molecules of the liquid species that do NOT move away from the surface may be reabsorbed into the liquid phase, thus reducing the rate of evaporation.
The two means by which molecules move away from the surface are convection (bulk flow) and diffusion. The greater the number of non-liquid-species molecules per volume, the lower the rate of diffusion.