Probably not. Given the same drive power, heat is determined by the efficiency of the emitter (radiometric W per electrical W), not efficacy (lm per W).
Even though 6500K is generally more efficacious than 4000K (more lumens per watt), in terms of efficiency, the difference will be much, much smaller. Most of the difference in emitter efficacy is explained by the greater spectral efficacy of the 6500K spectrum compared to the 4000K spectrum, not the actual electrical-radiometric efficiency of the emitter.