Suppose that $A$ and $B$ are two complete atomic Boolean algebras and $R$ is a relation between $A$ and $B$ with the following property:
If $Rab$ and $A^\prime$ is a finite Boolean subalgebra of $A$ containing $a$, there is a lattice embedding of $A^\prime$ into $B$ mapping $a$ to $b$ that is a subset of $R$.
I would like to know whether it follows that for any $a$ and $b$ with $Rab$, there is a lattice embedding of $A$ into $B$ mapping $a$ to $b$ that is a subset of $R$. (I suspect that this is false, but I cannot construct a counterexample.)
(By a lattice embedding I mean an injective function that preserves finite meets and joins.)
[In response to bof's answer below, I also want to stipulate that $A$ embeds into $B$.]
Assuming $2^{\aleph_0}=2^{\aleph_1}$ the following seems to be a counterexample.
Let $A=\mathcal P(\omega_1)$ and let $B=\mathcal P(\omega_0)$; then $A$ and $B$ are complete Boolean algebras with $|A|=|B|$. I'm not sure what definition of "order embedding" you're using, but I think the fact that $A$ contains a chain of order type $\omega_1$ while $B$ does not implies that there is no order embedding of $A$ into $B$. On the other hand, I think the relation $$R=\{(a,b)\in A\times B:b\text{ is an infinite and coinfinite subset of }\omega_1\}$$ satisfies your condition.
Instead of $|A|\le|B|$ maybe you want to stipulate that $A$ is order-embeddable in $B$? Or maybe you want to assume GCH?