The answer is yes.

The right ZFC definition of almost equivalent, in light of my comment, is that every condition in $\newcommand\P{\mathbb{P}}\P$ has a cone below that is forcing equivalent to a cone in $\newcommand\Q{\mathbb{Q}}\Q$ and vice versa. Without loss, we may take both $\P$ and $\Q$ as complete Boolean algebras.

The set of conditions $p\in\P$ whose lower cone $\P\newcommand\restrict{\upharpoonright}\restrict p$ is isomorphic to a cone $\Q\restrict q$ in $\Q$ is an open dense set in $\P$, and similarly with $\Q$.

So there is a maximal antichain $A_0\subseteq\P$ whose lower cones $\P\restrict a$ for $a\in A$ are each forcing equivalent to a cone $\Q\upharpoonright b$ for some $b\in B$. There is similarly an antichain $B_0\subseteq\Q$ whose cones $\Q\restrict b$ for $b\in B_0$ are each forcing equivalent to a cone $\P\restrict a$ for some $a$ refining an element of $A_0$. By iterating this process into the limit, I claim that we can find common refinements $A$ below $A_0$ and $B$ below $B_0$ such that every cone $\P\restrict a$ for $a\in A$ is forcing equivalent to a cone $\Q\restrict b$ for some $b\in B$ and vice versa.

In this case, we are done, since $\P$ will be forcing equivalent to the lottery sum of the cones $\P\upharpoonright a$ for $a\in A$ and similarly with $\Q$ and the cones $\Q\upharpoonright b$ for $b\in B$.

I had previously posted an answer, which was not correct.

The correct answer is negative, as shown in the first part of the answer of Calliope Ryan-Smith, following a suggestion of Andreas Leitz there in the comments.