Connecting 5 V at one end and GND at the other will dramatically reduce the differential between different LEDs, and improve the voltage at the lowest voltage LED ...
...assuming all the loads are drawing maximum current.
The resistance of the supply remains quite high though, and as the LED current drops, the voltage will rise. You will therefore still get a significant change in voltage and hence colour as you change the programmed brightness of the LEDs.
If you do have access to both ends, it's far better to supply 5 V and GND to both ends. This reduces the supply impedance and so the voltage swing with brightness significantly.
Here is a simple simulation of a power supply along a string of 4 loads drawing, in this example, 100 mA each. The supply has a 1Ω resistance wires between each load. The schematic shows the circuit in the 'alternate ends' configuration, drawing maximum current. With a very low programmed brightness, the voltage on all the loads will rise to 5 V.

Load
|
Single end
R1 5 V, R6 GND |
Alternate ends
R5 5 V, R6 GND |
Both ends
R1+5 5 V, R6+10 GND |
5 V one end Ground both ends R1 5 V, R6+10 GND |
I1 |
4.2 V |
3.6 V |
4.6 V |
4.4 V |
I2 |
3.6 V |
3.4 V |
4.4 V |
4.0 V |
I3 |
3.2 V |
3.4 V |
4.4 V |
3.8 V |
I4 |
3.0 V |
3.6 V |
4.6 V |
3.8 V |
Re-reading your question, it appears your diagram shows supply at one end, but ground connected to both ends. I've added this to the simulation. It's rather better than one end, and a bit better than alternate ends, but still not as good as using both ends.