This is a problem I am stuck at. First, I guess I have to use the maximum modulus principle to show that $I(r$) is strictly increasing. However $I(r)$ is defined by an integral, not some maximum value of $f(z)$, so I can't see how to apply the maximum modulus principle. Also, how can I find continuous $\varphi$ if $f(z)$ have zeros? Is φ guaranteed to be continuous at zeros of $f(z)$? If these mysteries are solved, then I think I can solve the exercise. Could please anyone help me?
Yes, $I$ is an integral; but so is $F$. Since $F(r)=2\pi I(r)$, the maximum principle applied to $F$ can lead to the result.
The continuity of $\varphi$ isn't really needed for the proof to work: $\varphi$ being measurable is enough to conclude that $F$ is a holomorphic function. Any reasonable integral involving a parameter in a holomorphic way is itself holomorphic, as can be verified using Morera's theorem (integrating over a triangle) and Fubini's theorem (exchanging the order of integration).
Also, the zeros of an analytic function form a discrete set. Hence, there is only a discrete set of values of $r$ for which $f$ has a zero on the circle $|z|=r$. Let's consider a circle without zeros. Then $\varphi(\theta) = |f(re^{i\theta})|/f(re^{i\theta})$ is continuous, and the function $$ F(z) = \int_0^{2\pi} f(ze^{i\theta})\varphi(\theta)\,d\theta $$ is holomorphic in $B(0;R)$. This function is constructed so that $F(r) = 2\pi I(r)$. By the maximum principle, $F(r)\le \max_{|z|=\rho} |F(z)|$ for any $\rho>r$. And since $$ \max_{|z|=\rho} |F(z)| \le \int_0^{2\pi} |f(\rho e^{i\theta})|\,d\theta = 2\pi I(\rho) $$ it follows that $I$ is increasing. In fact, strictly increasing, because nonconstant $f$ implies nonconstant $F$. Here's one way to argue this: if $F$ is constant, then $F(0)=F(r)=2\pi I(r)$. On the other hand, $F(0) = f(0)\int_{0}^{2\pi} \varphi(\theta)\,d\theta$. So, we must have $\varphi \equiv 1$, which means the imaginary part of $f$ is zero on the circle $|z|=r$. But $\operatorname{Im}f$ is harmonic and therefore obeys a maximum principle: it must vanish on a disk, and then $f$ is constant.
So we have $I(r)<I(\rho)$ for all pairs $(r,\rho)$ such that $r<\rho$, except for a discrete set of $r$-values. Since $I$ is continuous, the inequality holds in general; $I$ is strictly increasing.
Log-convexity is similar in principle, but I suppose one has to prove Hadamard's three circle theorem first.