Why use $\chi^2$ test to determine the presence of white noise?
If I and Q are normal, then their magnitude square is $\chi^2$ distributed. 

Hence, if you check the distribution of the instantaneous power against that distribution, you get information about how much the complex baseband signal is circularly complex normal.

The autocorrelation function has power of the signal as its value for zero shift. For whiteness, the sample autocorrelation approaches an accordingly weighted Dirac impulse with growing observation length.