To be sure there is no conflation between "phase" and "delay" in the time domain, I refer to a phase shift in time as:
$$y(t) = x(t)e^{j\theta}$$
Whereas a time shift (delay) is given as:
$$y(t) = x(t-\tau)$$
As detailed below they are related, but not the same. The answer below uses "phase" and "time offset" as represented above, and in the context of GPS C/A codes and PRN codes at complex baseband where correlation would typically be applied (not passband signals).
A phase in the cross-correlation of two CDMA Gold Code sequences or PRN sequences indicates a frequency offset, not a time offset of the baseband signal. If the two sequences are real, with no frequency offset, the autocorrelation function (which is the correlation for each delay between the sequences) will be real.
For small frequency offsets, the phase change of the correlation between a PRN sequence and a frequency shifted copy of itself at 0 delay between the two is approximately given by:
$$\Delta\phi = -2\pi f_o T \tag{1} \label{1}$$
Where $\phi$ is the change in phase in radians,
$f_o$ is the frequency offset in Hz, and
$T$ is the time duration of the correlation in seconds.
Where "small frequency offset" is $f_o << 1/T$ (the correlation magnitude goes to 0 at $f_o=1/T$, and then continues as a Sinc function with frequency offset.)
(Also as a side to note of possible interest, as @vml points out in the comments, this is directly related to the coherence time, as the correlation time duration $T$ gets longer the amount of tolerable frequency offset for the signals to still be correlated gets smaller; for more details on that see this post and it's linked post).
The sign is negative vs frequency offset assuming the correlation is computed according to:
$$\rho = \int_0^T x(t) g^*(t) dt\tag{2} \label{2}$$
With $g(t)$ as the frequency shifted version of $x(t)$ as $g(t)=x(t)e^{j2\pi f_o t}$.
The relationship given in \ref{1} is understood from the correlation formula and the Fourier relationships for frequency and time offsets:
A delay (time offset) in time will be a phase in the frequency domain, given the Fourier relationship:
$$x(t-\tau) \leftrightarrow X(f)e^{j2\pi f \tau}$$
This relationship also explains how the phase is unaffected by the time delay a baseband signal with no carrier frequency offset (when $f=0$), while a static phase would be introduced for the case of a passband signal where $f \ne 0$.
An similarly an offset in frequency will be a phase in the time domain:
$$x(t)e^{j2 \pi f_o t} \leftrightarrow X(f-f_o)$$
Which would then appear in the autocorrelation function as a phase offset between the C/A code and the same C/A code with a frequency offset using the relationship above.
From \ref{2} consider a case when the waveform is correlated to itself with a frequency offset starting from a condition where the phase of the two is zero at $t=0$, the peak correlation is given as:
$$\rho = \int_0^T x(t) x^*(t)e^{-j2\pi f_o t} dt\tag{3} \label{3}$$
The conjugate product $x(t)x^*(t)$ will always be real, and thus zero phase. Therefore the phase of $\rho$ is the result of integrating over the complex trajectory given by $e^{-j2\pi f_o t}$. This phase of this trajectory proceeds linearly from $0$ radians to $-2\pi f_o T$ radians, and thus the phase of the resulting integral of a phase ramp will be $(-2\pi f_o T)/2 = -\pi f_o T$.
For the $\Delta \phi$ as given above, which is the change in phase from one correlation interval to the next (assuming we did block by block correlations over time $T$), the phase trajectory would go from $-2\pi f_o T$ to $-4\pi f_o T$ over the next correlation duration (integrating from $T$ to $2T$), with the resulting phase $-3\pi f_o T$ (and thus the change in phase is as given in $\ref{1}$.) Also if we were to do the correlation over a moving window $T$ long, the phase that we observe versus time would be consistent with the frequency offset as frequency is a change in phase over a change in time (time derivative of phase.
Confirming this is a plot below showing the real and imaginary output of the correlator for GPS C/A SV24:
Repeating the above as a cross-correlation of the CA-Code for SV24 with the same but having a 20 Hz frequency offset results in the an imaginary component for the correlation (Note, this is a frequency offset in the signal itself, not the sampling clock. When the sampled signal is not exactly at baseband meaning zero frequency offset, an offset in the sampling clock would also modify that frequency offset in the signal, as a ratio proportional to the frequency offset of the signal over the sampling rate - I go into more details of the relationship between the two in my answer to DSP.SE #67990):
The amplitude of the real component is 1020.3 and the imaginary component is -64.2, thus the resulting angle is $\tan^{-1}(-64.2/1020.3) =0.0628$ radians. The correlation time $T$ was over 1 C/A code sequence or 1 mS, so the predicted angle given by \ref{1} with $f_o=20$ Hz is:
$$\phi = -\pi f_o T = \pi (20)(1e-3) = 0.063 \text{ radians}$$
Note how for small frequency offsets how the imaginary portion of the correlator output will be directly proportional to the frequency offset, and thus can be used as a frequency discriminator in a carrier recovery loop.
Continuing the above example where the correlation over a time duration $T$ of 1 C/A code sequence (1 ms) to a received GPS signal with 4 consecutive C/A code sequences (4 ms of a GPS capture), with and without the 20 Hz frequency offset condition, starting from the condition of no carrier phase offset:
The above plot is the real and imaginary components of the correlator output for the 4 consecutive C/A code sequences with a carrier offset, correlated against the reference code in the receiver. If we plotted that same result on a complex IQ plane, we can see how the phase rotation with $\Delta \phi$ between each result is consistent with the explanation provided above (we can measure precise carrier phase and frequency offsets from the correlator result, given a change in carrier phase versus time IS frequency offset, while time offset at baseband is the result we see when we shift right or left along the time axis given above):
