Photodetector Question: Does converting an RGB image to grayscale produce the same result as using a grayscale image detector? FYI my application is not astronomy related, but I know astronomy folks have a lot of knowledge of photodetection and spectrometry instrumentation, so I am seeking this community's help. 

I'm interested in doing both spectral and RGB imaging. I'm wondering if I can do both with just an RGB CMOS camera... The way I would approach the spectral imaging is to illuminate my sample with monochromatic light one wavelength at a time and then convert the images to grayscale. Each grayscale image would represent the intensity of light reflecting from my sample at a particular wavelength. I would love for this to work, but I am doubtful that the converted image will actually represent what a grayscale camera would record. 

Bit of an odd scenario... Wasn't sure how to search the web for this. Any help is appreciated! Thanks!
Really, all cameras/detectors are greyscale detectors.  Every pixel in a CCD detector puts out a single number for the number of photons it detected. To create an RGB image you need three filters.  If you have 3 detectors, you can put a different filter in front of each and get 3 numbers at once.  If you have 1 detector, then you take 3 measurements with a different filter each time.  To create the color image, one combines these values but you have freedom to weight them as you like.  Typically, one tries to imitate the sensitivity of the eye, but that is a complicated issue since the sensitivity of the eye to different colors changes with total intensity and the filters can not exactly reproduce the eye's bandpasses.  So, there is no right way to do this.  Some people like red more than blue, so they weight the R-band more. Once, you get aR+bG+cB, you loose the information you need to derive R, G, and B separately.  

So, to do your experiment properly, you need to first get the response function (response vs wavelength) of the filters + CCD of your camera at each wavelength that you are using.  You do this by imaging without the sample.  Then repeat with the sample and divide "with" by "without", where "with" is the output in the R, G, or B channel that has the most response in the "without" case.