@YCor. I have done it for all non-abelian groups of exponent $3$ of order at most $3^6$. There is no such example among such groups.

\begin{align*}‎ ‎|\{(x,y)\in L\times L \;|\; [x,y]=0\}|&=&\\‎ ‎&\sum_{x\in L} |C_L(x)|= \sum_{i=1}^{n} |C_L(x_i)||[x_i]|\\‎ ‎&=\sum_{i=1}^n |C_L(x_i)||L:C_L(x)|=\sum_{i=1}^n |L|=n|L|‎. ‎\end{align*}‎ ‎Hence the ratio is equal to $n$‎. ‎This completes the proof‎.

(2) $\sim$ is transitive‎. ‎Let $x\sim y$ and $y\sim z$‎. ‎Then there exist $t,s\in L$ such that $x+[x,t]=y$ and $y+[y,s]=z$‎. ‎It follows that‎ ‎\begin{align*}z=y+[y,s]&=&\\&x+[x,t]+[x+[x,t],s]=\\‎ ‎&x+[x,t]+[x,s]+[x,t,s]=x+[x,t+s]‎. ‎\end{align*}‎ ‎Thus $z=x+[x,t+s]$ and so $x\sim z$‎. ‎ ‎Thus‎, ‎$L$ is partitioned by $[x]$'s‎. ‎Suppose that $[x_1],\dots,[x_n]$ are all orbits of the equivalence relation $\sim$ on $L$‎. ‎Now note that is any Lie ring‎, ‎we have $|L:C_L(x)|=|[x]|$‎. ‎Therefore‎,

$$x \sim y \Leftrightarrow \exists\; t\in L \;\;\;\; x+[x,t]=y.$$‎ ‎We now prove that $\sim$ is reflexive‎, ‎symmetric and transitive‎. ‎ ‎(1) $x+[x,0]=x+0=x$ ($\sim$ is reflexive)‎. ‎\item If $x\sim y$‎, ‎then there exists $t\in L$ such that $x+[x,t]=y$‎. ‎Thus $[y,t]=[x+[x,t],t]=[x,t]+[x,t,t]$ which implies $[y,t]=[x,t]$‎. ‎So‎, ‎$x=y-[x,t]=y-[y,t]=y+[y,-t]$‎. ‎Thus $y\sim x$‎. ‎Hence $\sim$ is symmetric‎.

‎Let $L$ be a Lie ring such that $[L,L,L]=0$‎. ‎Then $c(L)$ is an integer‎. ‎ ‎For amt $x\in L$‎, ‎define $[x]=\{x+[x,y] \;|\; y\in L\}$‎. ‎The latter set is an orbit of the equivalence relation $\sim$ on $L$ which is defined as follows‎:

@YCor. I used "always" in the title so I have to write in positive. I wrote the question in the body in negative as my guess is that an example is elusive.

@DavidSpeyer. I have a proof for a finite nilpotent Lie ring of nilpotent class $2$. I will publish it here tommorow, as it is now in my computer at university.

@DavidSpeyer. In the meanwhile, the nilpotency class of a possible counterexample should be at least $3$. I have proved the latter.

@DavidSpeyer. Yes. By Lazard's correspondence there is a bijective map $\mathcal{L}$ from an arbitrary nilpotent finite dimensional Lie algebra $L$ over a finite field of characteristic $p$ of nilpotent class less than $p$ to a finite $p$-group $G$ (of the same nilpotency class as $L$) such that $[x, y]^\mathcal{L} = [x^{\mathcal{L}}, y^{\mathcal{L}}]$ for all $x, y \in L$, where the right hand side is the usual group commutator $[a, b] = a^{-1}b^{-1}ab$. This implies that the answer to the question is negative for all such Lie algebras $L$.

@YCor. I think I have missed another hypothesis that the Lie algebra should be nilpotent.

@YCor: Could you please kindly write down your groups $G$ and $H$ in details. It is very interesting for me too. But I could not follow your quick point in the above.

@M.FarrokhiD.G.: My proof depends on calculation and seemingly is ``$p$" sensitive.