Let $f$ be a non-vanishing holomorphic on the unit disk $D$. Suppose $|f|$ converges to a measure $\mu$ on $\partial D$ as $|z|\rightarrow 1$, in the sense that $$ \int_{\partial D} |f(r z)| \phi(z) dz \rightarrow \int_{\partial D} \phi(z) d\mu(z) $$ as $r\rightarrow 1$ for each continuous function $\phi$ on $\partial D$ (here the first integral is with respect to uniform measure on $\partial D$). If $\tilde f$ is another non-vanishing holomorphic function such that $|\tilde f|$ converges to the same measure $\mu$, must it hold that $f = \tilde f$?
I know that this holds if the boundary data for $|f|$ is continuous, as in this case $\log|\tilde f / f|$ is a harmonic function which vanishes on the boundary, so must be identically 0. However, I am not sure how to generalize the result to measure-valued boundary data.
I do not need a full proof---a reference indicating that this is true of false would suffice.
Presumable, you meant $f=\tilde f$ up to a unimodular factor. The answer is still no, however. Let $f$ be a singular inner function. Then $|f(rz)|\to 1$ a.e. as $r\to 1^-$, and $|f|\le 1$ everywhere. By the dominated convergence theorem, $|f |\to 1$ in the sense of $L^1$ convergence, and a fortiori as a measure. Thus, $ f$ shares the boundary values of $|f|$ with $\tilde f(z)\equiv 1$. (And with all other singular inner functions.)
You may be interested in the inner-outer factorization.