I strongly suspect that the choice of bulb will make no noticeable difference in evaporation rate. The total energy density striking the floor due to a 100 W bulb, even assuming that all of the energy is converted to light, is on the order of 120 $\mu \mbox{W}/\mbox{cm}^2$, which is tiny, and real bulbs are only on the order of 1% to 5% efficient, with most of the rest of the energy carried away as heat via convective cooling of the lamp surface by air. So, it's probably more like at most 20 $\mu \mbox{W}/\mbox{cm}^2$, and in either case, this is a tiny power density.
In contrast, you need power densities on the order of 10,000 $\mu \mbox{W}/\mbox{cm}^2$ to be able to see radiation make a noticeable color change on thermal-sensitive liquid crystal sheets, and those are thin, unlike the floor, which is thick and thus can also dissipate heat via conductive cooling. So the radiation due to a single light bulb is unlikely to have any impact on the heating and subsequent evaporation of water on a wet floor.
Likewise, the amount of light emitted by each bulb which is on-resonance with water transitions (microwave and infrared absorption lines) is probably not relevant. This is because in the condensed phase, rotational and vibrational relaxation times (which redistribute energy from the excited states into thermal energy) are on the order of milliseconds or shorter, whereas the evaporation process takes on the order of days or longer, so the system will more or less be at thermal equilibrium. So while a few water molecules may every now and then get excited, they'll hop back down immediately and shed the energy as heat, so it probably won't be any different than the nonresonant heating.
A large room fan blowing air around, however, may speed things up noticeably.
There are several ways to reduce the rate of evaporation of water. However, first it is important to understand the factors influencing the rate of evaporation:
1) Amount of humidity in the air
2) Speed of air flowing past the water droplets
3) Intramolecular bonding strength
4) Pressure
There are more such as surface area, density, etc. but since we are considering the same drop of same substance these can be ignored. (temperature is there too, obviously)
So what we can do:
Make the air around the drops more humid.
Put the drops in a closed environment, to reduce the speed of air flowing past it. For example, water droplets inside a room will evaporate slower than water droplets placed outside.
There certainly are some additives to help increase the intramolecular bonding strength of a substance. Also, the substance the water droplets are placed on also makes a difference on the rate of evaporation.
Decreasing your altitude will also decrease the rate of evaporation.
Best Answer
What makes water boil/evaporate is the thermodynamic concept derived from the first and second law of thermodynamics.
You can read this article to find out the derivation from entropy to the Clapyeron equation.
http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation#Derivation_from_state_postulate
$$\frac{\mathrm{d} P}{\mathrm{d} T} = \frac {\Delta{H_{vap}}}{T \Delta v}.$$
And the integral form is:
$$\ln P = -\frac{\Delta{H_{vap}}}{R}\left(\frac{1}{T}\right)+C.$$
where $P$ is the vapor pressure at that particular temperature.
http://en.wikipedia.org/wiki/Vapor_pressure
For example, at room temperature (20C), the vapor pressure of water is about 2.3kPa. This means that water will evaporate until it reaches a partial pressure of 2.3kPa and then the water will attain 100% relative humidity (which means no more water can evaporate).
The driving force of this evaporation is thermodynamic related. Evaporation occurs in order for the system to reach thermodynamic equilibrium.
Below is a phase diagram - the water-water vapor phase boundary is simply the plot of vapor pressure against temperature.