This is not the case. Here is a counterexample in the case where $n=2$.
Suppose that $A$ is a countable dense subset of $\mathbb{R}^{2}=\mathbb{C}$. Let $r_{a}>0$ for each $a\in A$, and suppose that $\sum_{a\in A}r_{a}\cdot(1+\max(0,\log(a)))<\infty$.
Let $p(z)=\sum_{a\in A}r_{a}\cdot\log(|z-a|)$. Let $s:\mathbb{C}\rightarrow(0,\infty)$ be any subharmonic function. Let $q(z)=p(z)+s(z)$. Then observe that $p(z)=q(z)=-\infty$ on a dense $G_{\delta}$-set. In particular, $p=q$ on a dense set. The functions $p,q$ are not locally bounded, but we can obtain locally bounded functions from $p,q$.
The function $p$ cannot be $-\infty$ everywhere since
$$\frac{1}{2\pi}\int_{0}^{2\pi}p(re^{i\theta}+z_{0})d\theta=(\sum_{a\in A,|z_{0}-a|<r}r_{a})\cdot\log(r)+\sum_{a\in A,|z_{0}-a|>r}r_{a}\cdot\log(|z_{0}-a|)>-\infty.$$
In fact, the set $p^{-1}[\{-\infty\}]$ is a polar set and polar subsets of $\mathbb{C}$ have Hausdorff dimension $0$ and therefore Lebesgue measure zero as well.
Let $u(z)=e^{p(z)},v(z)=e^{q(z)}$. As a consequence of Jensen's inequality, the functions $u(z),v(z)$ are subharmonic, locally bounded, and $u\leq v$. However, $u(z)=v(z)=0$ on a dense $G_{\delta}$-set, and if $p(z)>-\infty$, then $p(z)<q(z)$, so $u(z)<v(z)$.
The fact that $u(z)=v(z)=0$ on a dense $G_{\delta}$-set should be contrasted with the fact that if $f,g$ are subharmonic functions that are equal almost everywhere, then $f=g$.
One way to derive the global inequality is the Principle of Harmonic Majorant: If $D$ is a bounded region, $u$ is harmonic
in $\overline{D}$, $f$ is analytic in $\overline{D}$, then the inequality $\log|f(z)|\leq u(z),\, z\in\partial D$ implies the same in $D$. To prove this, just apply the Maximum principle to the harmonic function
$\log|f|-u$ in the region $D$ minus small neighborhoods of zeros of $f$; that the inequality holds on the boundaries of these small neighborhoods is clear since $\log|z|\to-\infty$
when $z\to$ a zero of $f$.
Now to prove your global inequality, use as $u$ the solution of the Dirichlet problem with the boundary values $\log|f|$.
The solution of the Dirichlet problem for a disk is completely elementary, and the value at the center is just the average of the values on the circumference.
So we only used the Maximum Principle for harmonic functions and solvability of the Dirichlet problem for a disk.
Best Answer
Conditions for a log-subharmonic function $f$ on $D\in\mathbb{R}^n$ are described by Mochizuki in A Class of Subharmonic Functions and Integral Inequalities (2004).
The conditions are phrased in terms of an inequality for the volume average $A_p$ of $|f|^p$ and the surface average $M$ of $|f|$, in the form $A_p\leq M^p$ for any closed ball in $D$.
If $n\geq 2$ the condition $A_{1+2/n}\leq M^{1+2/n}$ is sufficient for $\log|f|$ to be subharmonic. If $n=2$ this condition is also necessary, if $n>2$ it is not.