It is a very nice question!
Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.
Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED
I expect that a natural generalization of the argument will apply at higher cardinals.