In general it is impossible. We can take the square root, map the circle conformally to the half-plane, and arrive at the following problem: given any nonnegative $f\in L^2(\mu)$ with finite logarithmic integral, can we find $g\in H^2(\mu)$ with $\Re g\ge f$ and $|\Im g|\le \Re g$ where $d\mu(x)=\frac{dx}{1+x^2}$? Now, if that is possible for all $f$, it is also possible with the $H^2(\mu)$ norm of $g$ bounded by $C\|f\|_{L^2(\mu)}$.
Let's do the usual blow-up now taking some $f\in L^2(dx)$ and considering $f(nx)$ instead of $f$. Then, when we scale back, we'll get majorants $g_n\in H^2(\mu_n)$ such that $\|g_n\|_{H^2(\mu_n)}\le \|f\|_{L^2(\mu_n)}\le \|f\|_{L^2(dx)}$$\|g_n\|_{H^2(\mu_n)}\le C\|f\|_{L^2(\mu_n)}\le C\|f\|_{L^2(dx)}$ with $d\mu_n(x)=\frac{dx}{1+(x/n)^2}$. We can pass to a subsequence and assume that $g_n$ converge to some $g$ weakly in $L^2(dx)$ on every subinterval of $\mathbb R$. Then we shall have $g\in H^2(dx)$ and we still have $\Re g\ge f, |\Im g|\le \Re g$(the vanishing of the Fourier transform on the negative semi-axis and the comparisons with non-negative functions can be tested by integrating against appropriately chosen $L^2(dx)$ functions, and the norm can only drop).
But for $H^2(dx)$ functions we have $\int_{\mathbb R}|\Re g|^2dx=\int_{\mathbb R}|\Im g|^2$, so we are forced to have $|\Im g|=\Re g$ almost everywhere on the line. But then $g^2\in H^1(dx)$ and $\Re[g^2]=0$ on $\mathbb R$, which is impossible.
Thus, a sufficiently strong bump at one point will give you a counterexample. To figure out exactly how strong is "sufficiently strong", one needs to make all that weak limit nonsense quantitative, which I leave to someone else :-)