Here is a simple answer in a special case when $\phi$ is harmonic.

** Geometric proof:** Let $g_0$ be the (incomplete) metric on the unit disc induced from $\mathbb R^2$. Then the metric $e^\phi g_0$ is complete if and only if the integral
of $e^\phi$ over $\gamma$ is infinite for each $C^1$-curve $\gamma$ from the center
to the boundary of the disk. Now the sectional curvature of $e^\phi g_0$ is $-\frac{1}{2}e^{-\phi}\Delta\phi$, which is zero since $\phi$ is assumed harmonic. Thus if $e^\phi g_0$ were complete, it would be isometric to the standard $\mathbb R^2$, but isometries are conformal, so we would get that the unit disk is conformal to the plane.

** Sketch of complex analysis proof:** If by a "direct proof" you mean then one that uses only complex analysis, then Huber's proof quoted in my question does just that. However,
Huber's argument simplifies a lot when $\phi$ is harmonic. Indeed, harmonicity of $\phi$ allows you to find its harmonic conjugate, and since the disk is simply connected, $\phi$ becomes the real part of an analytic function $\tau$ on the disk.
Thus $e^\phi=|e^\tau|$, and $e^\tau$ is analytic and nowhere vanishing. Now I think the simple argument in the note "Paths of rapid growth of entire functions" by Kaplan does the job (see middle of the first page).

** Remark**. Kaplan's note is in fact an elaboration of the Huber's proof, and it gives a simple answer to your question for all superharmonic functions of the form $\phi=-\log|f|$ such that $f$ does not vanish (and this is what you seem to care about by restricting to smooth superharmonic functions). For general superharmonic functions I do not see how to do better than Huber.