Dropping voltage for red channel on RGB LED
LEDs need to be driven via current regulation rather than voltage regulation. Despite the G and B color channels being rated at 3 to 4 Volts, providing them with a 3.3 Volt regulated input without a current limiting resistor may to cause the LED to burn out.

One possible (but not practical in this case) solution is to use 3 resistors on the 3 driven legs of the LED (not just one resistor on a combined current return path). Calculate the resistance according to the current desired for each color, and the rated minimum forward voltage for each, e.g. :


  R  = (Vsupply - Vfwd) / I
  
  Rr = (3.3 - 1.8) / 0.35  = 4.28 Ohms
   Rg = Rb = (3.3 - 3.0) / 0.35 = 0.85 Ohms
     (assuming 350 mA drive current, 3.3 Volt regulated input, no other voltage drops)


You probably will not want to look for 0.85 Ohm resistors for Green and Blue. Also the ~4.3 Ohm resistor for Red will need to be rated at 1 Watt, since the computed dissipation for it works out to 0.525 Watts. Hence this isn't a practical approach.

Please note that if these LED channels are to be switched or controlled, rather than permanently on, your switching solution (MOSFET, transistor, or something else) will introduce a voltage drop on each color's current path. This complicates the calculation further, especially when working with the 0.3 Volt headroom the proposed 3.3 Volt regulator will provide.

It would be better to use a higher voltage to drive the LED channels, dropping it down suitably with resistors, or preferably using current regulators such as SD42351 or one of the SuperTex LED driver products. 

If using resistors, you need two of value Rg and one of value Rr calculated for Vsupply of 5 Volts, and wattage rating higher than the calculated power dissipation for each resistor ( = Vdrop * 350mA).