This is the circuit you are proposing: -
I've put a circle in red around the 1 ohm resistor because this is important about your concept. To activate the TIP2955 you'll need at least 0.7V across the 1R resistor and now your attempts to make a LDO circuit are reduced because.
With 0.1A flowing, the regulator drop out is added to 0.1V across the resistor
With 0.5A flowing, the regulator drop out is added to 0.5V across the resistor
Have you taken this into account?
EDIT - following advice offered on an edit suggestion (which somehow disappeared) I'll add to the scenario. What I said above still stands but as the demand for more load current increases, so does the base emitter voltage on the TIP2955 (at 6A this may be in the order of 1.4V). And, for a 6A load, it is likely that the Hfe of the transistor will be about 30 (typical) implying the base current will need to be about 200mA. The 10R base resistor will need to pass this current and therefore, in adddition to the 1.4V required by the base there will be 2V across the 10R resistor - this equals 3.4V - all required to be generated across the 1R series resistor. This implies a 1R current of 3.4A and of course this is not viable through the regulator chip (1.5A max). In short this circuit will never be able to provide a 6A drive to the load.
If you got rid of the 10R resistor and replaced it with a short, you'll still need to produce 1.4V across the 1R and this just about means the circuit will work with 6A BUT, it's on the edge and there has to be a better way of doing this. The CE saturation voltage is not that good at 6A either so I'd abandon this idea and look for alternatives.