Dual voltage LED circuit
No, it doesn't work that way.  The two LED strings are in parallel.  That means the both see the 3.3V independently.  Let's say the LEDs drop 2.0V, which appears is what you are assuming from your numbers.  That means when SRC is 5V, there will be 3V on R1.  3V / 150&Omega; = 20mA as you said.  However, when SRC is 3.3V, the same logic still applies.  The LED will still take about 2V, so there will be 3.3V - 2V = 1.3V accross R1.  1.3V / 150&Omega; = 8.7 mA, which will be the current thru D1 when SRC is 3.3V.

If this extra 9 mA is of no consequence to your 3.3V supply, then you can simply leave R1 and D1 on the board.  If the 9 mA matters, then you have to do something.  Removing either R1 or D1 would do it.

Another thing to consider is that you don't need 20 mA thru D1 when debugging.  You probably got that from the datasheet, which shows 20 mA as the maximum.  That's a common value for small LEDs.  However, it will still be plenty bright enough to see on your bench at 5mA.  Since R1 is dropping 3V, it can be 600&Omega; to allow 5 mA to flow.  Then at 3.3V you only get 2.2 mA thru R1 and D1.  That's a lot less than 9 mA as you originally had it.  If 2.2 mA is tolerable, then you need to do nothing more.