You can get a robust version for free by a precompactness argument: e.g., if $\|f\|,\|g\|\leq 1$, then for any $\epsilon>0$, there exists $\delta>0$ such that $\||f|-|g|\|_{\mathbb{D}}<\delta$ implies $\|f-cg\|_{(1-\epsilon)\mathbb{D}}<\epsilon$ for some $c$. Here the norm may be $L^\infty$ or $L^2$-norm or $H_2$-norm.
Assume the contrary; then there are sequences $f_i,g_i$ and an $\epsilon>0$ such that $$ \||f_i|-|g_i|\|\to 0;\quad \|f_i-cg_i\|_{(1-\epsilon)\mathbb{D}}>\epsilon\quad\forall c. $$ Norm-boundedness for analytic function implies uniform boundedness on compact subsets of $\mathbb{D}$, so by passing to a subsequence we may assume that $f_i\to f$ and $g_i\to g$ uniformly on compact subsets of $\mathbb{D}$. But then $|f|\equiv|g|$ but $\|f_i-cg_i\|_{(1-\epsilon)\mathbb{D}}\geq\epsilon$ for all $c$, a contradiction.
