I see a fundamental units problem in your math. You are using the symbol m which stands for milli (10^-3). The actual energies are µ (10^-6). You are off by a factor of 10^-3. Since most keyboards do not have a symbol for µ (micro), people often revert to an m (milli), and end up confusing the units. Hey, they both start with the same letter, right? :)
Additionally, there is the question of "wall plug efficiency". While a diode may 60mW, even military grade lasers only have a 10-15% wall plug efficiency. Thus that means an output of around 6mW. Then any optical element will reduce power by around 50% for each element. Assuming at most 2 optical elements (nu-naturally low number), that means the output can at max be 1.5mW.
In the answer you quoted, use this paragraph as a point of comparison:
For the sake of comparison, sunlight is one kilowatt per square meter and perhaps 5% of that is near infrared i.e. 700 to 1000 nanometers. Just going outside will expose you to much greater power densities of SWIR than the Kinect.
Also, remember that even though the generator is 60 mW (yes, I used the correct units), there is a series of diffusers, optics, and such so that at the very extreme of the exit aperture, the power density is <25 μW (again, note the symbol). The series of steps required to get at the 60 mW generator would indicate a willful intent to cause self harm, and be beyond simple mechanical failure.
Your initial assumption is incorrect.
My approach is as follows: - Assume 60mW output power is correct - Diffuser efficiency is 50% and therefore 50% of the energy is lost
The diffusers and optics reduce the power to <25 μW at the aperture. Run your math with that figure and you'll have an accurate representation.
In general, you can detect laser with solar panels. There is intensive research on possibilities of employing lasers in wireless power transmission (Laser Power Beaming). Laser Power Beaming have already been used in order to provide energy for remotely controlled machines (example).
It is my belief that we are 20-30 years from the point when Laser Power Beaming from orbital based photovoltaic power collectors (satellites) will be possible (remember SimCity?).
That said, it seems like you are not that into wireless power transmission. I guess you want to use laser as means of transmitting some information to an electrical circuits which get attached to humans clothes, right?
Well, I've seen (and used) few military combat simulation systems which employ lasers in order to allow soldiers to shoot at each other (without killing each other). Even those systems which are vehicle mounted did not use photovoltaic cells in order to detect laser radiation.
We could provide you with a bunch of information about photovoltaics and lasers, but I believe that your initial direction is incorrect. You can use several photodiodes scattered over the body in order to increase the coverage of your system - it will be cheaper, simpler and much more durable approach.
If sensitivity is your concern, you might want to check out for photomultiplier. These devices have very high sensitivity, but are more complex and more expensive than photodiodes. They are also much more prone to mechanical damage.
Best Answer
What is typical is to use a narrow band wavelength filter to knock out all but the laser wavelength. But keep in mind that the bandpass of this filter will be angle dependant.