The warm white phosphor dissipates more energy at a given power level than a cool white phosphor at the same current.
For example:
Say you have a die that is a Q5 (107 lm min. @350mA) under a cool white phosphor but is a P4 (80.6 lm min. @350mA) under a warm white phosphor. Let's also say that the LED has a Vf of 3.50V at 350mA. The same amount of energy enters each phosphor, but different amounts are dissipated as light. For the sake of comparison, we'll use 242.5lm/W as to convert light output to power. The real number shouldn't be too far from that.
At 350mA, the warm white phosphor dissipates 26.4 lumens' worth of power more than the cool white phosphor, or about 109mW more than the cool white phosphor.
If we say that the bare blue die has a luminous efficacy of 140lm/W, then the cool white phosphor dissipates the difference between the input power to the phosphor (140lm/W*1.225W) and the output power dissipated by the phosphor (107lm / 242.5lm/W), or more concisely, 707mW - 441mW, or 266mW dissipated as heat by the phosphor.
For warm white, the efficacy of the blue die remains the same, but the efficacy of the phosphor decreases. The numbers are 707mW - 332mW, or 375mW dissipated as heat by the phosphor.
The warm white phosphor dissipates as heat 29.0% more power than the cool white. What I'm getting at is that the phosphor of a warm white LED will settle at the same temperature as a cool white LED running at 29% higher power. The phosphor of a warm white LED running at 700mA will have roughly the same temperature as that of a cool white LED running at 1000mA.
So, If you are planning to overdrive a warm white LED, the phosphor will have either approximately 30% less lifetime than a cool white LED at the same current or will have approximately the same lifetime driven at 30% lower power consumption.