This is the isoperimetric inequality in disguise. Define the Riemannian metric
with length element $e^{u(z)}|dz|$. The curvature of this metric is
$$-e^{-2u}\Delta u=0,$$
so the metric is flat. Then your inequality says that the area of the disk is
at most $1/4\pi$ times the square of the length of the boundary, the usual isoperimetric inequality for the flat metric.

Remark. It is sufficient to assume that curvature is non-positive, that is $\Delta u\geq 0$.

Ref. MR0936419 Burago, Yu. D.; Zalgaller, V. A. Geometric inequalities. Translated from the Russian by A. B. Sosinskiĭ. Springer-Verlag, Berlin, 1988.

Isoperimetric inequality for arbitrary surfaces is usually called Fiala-Aleksandrov inequality. The original publications are:

MR0006422 Fiala, F. Le problème des isopérimètres sur les surfaces ouvertes à courbure positive, Comment. Math. Helv. 13 (1941), 293–346, and

MR0052136 Aleksandrov, A. D. An isoperimetric problem, Doklady Akad. Nauk SSSR (N.S.) 50, (1945). 31–34.