Answer using Three Circles and Maximum principle:
For any $r>0$, let $m(r) = \sup_{|z| = r} u(z)$. Fix $0. The three circles theorem says that $m(b)\leq \frac{\log r - \log b}{\log r - \log a}m(a) + \frac{\log b - \log a}{\log r - \log a}m(r).$ Taking the $\limsup$ as $r\to \infty$ of both sides of this inequality, I get that $m(b)\leq m(a)$, since by your assumption $\limsup_{r\to\infty} m(r)/\log(r) = 0$. By the maximum principle, we conclude that $m(a) = m(b)$, and that $u$ is constant on $\{|z|\leq b\}$. But $a$ and $b$ were arbitrary, so $u$ is constant on $\mathbb{C}$.
Answer using Jensen's formula:
Jensen's formula for subharmonic functions says the following.
Let $u\colon \mathbb{C}\to \mathbb{R}\cup\{-\infty\}$ be a subharmonic function such that $u(0) \neq -\infty$. Let $\mu = \Delta u$ be the Laplacian of $u$ (so it is a positive Radon measure on $\mathbb{C}$). Then $\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta = u(0) + \frac{1}{2\pi}\int_0^r\frac{\mu(\mathbb{D}_t)}{t}\,dt,$ where $\mathbb{D}_t$ is the open disk of radius $t$ around the origin.
Let $u$ be a subharmonic function satisfying your $\limsup$ condition. Up to translating it, you may assume $u(0) \neq -\infty$. Unless $u$ is harmonic, the measure $\mu = \Delta u$ is nonzero, and hence there is some radius $R>0$ for which $\mu(\mathbb{D}_R)>0$. It follows from Jensen's formula that for $r>R$ $\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta =u(0) + \frac{1}{2\pi}\int_0^r\frac{\mu(\mathbb{D}_t)}{t}\,dt\geq u(0) + \frac{\mu(\mathbb{D}_R)}{2\pi}\int_R^r\frac{dt}{t}$$ = u(0) + \frac{\mu(\mathbb{D}_R)}{2\pi}\log(r/R)\geq \mathrm{const}\cdot\log r.$ On the other hand, because of your assumption on $u$, for any $M>0$ one has $\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta = \frac{1}{2\pi}\int_0^{2\pi}\log(r)\frac{u(re^{i\theta})}{\log r}\,d\theta\leq M\log r$ when $r\gg 1$. These two inequalities give a contradiction. It follows that $u$ must be a harmonic function. Dealing with this case should be easier.