Let $A$ be a unital nonzero Boolean algebra of cardinal $\ge 4$. Then $A$ has a unital Boolean subalgebra of index 2. (This excludes being Jónsson at all cardinals.)

Indeed, $A$ has at least two (unital ring) distinct homomorphisms onto $\mathbf{Z}/2\mathbf{Z}$. This gives rise to a surjective (unital ring) homomorphism onto $(\mathbf{Z}/2\mathbf{Z})^2$. The inverse image of the diagonal yields the desired subalgebra.

Topologically, this corresponds to taking two points in the Stone space and gluing them.

(Edited Oct 19, 2018)

Actually, no scalar (=associative unital commutative) ring is Jónsson. Here by Jónsson I mean: uncountable and every proper unital subring has smaller cardinal.

First, a domain $A$ is never Jónsson (among unital rings). Indeed, we can consider a maximal algebraically free subset $X$ of $A$; since $A$ is uncountable, we have $|X|=|A|$, and hence we can construct, proper unital subrings of the same cardinal as $A$.

For $p$ prime or zero, say that a scalar ring is $p$-reduced if it embeds into a product of domains of characteristic $p$. Then an uncountable $p$-reduced scalar ring $A$ is never Jónsson. Indeed, write $A\subset\prod A_i$, with $A_i$ domain of characteristic $p$. If some projection has cardinal $|A|$, uncountable, we can pull back some proper subdomain of cardinal $|A|$. Otherwise, each projection has cardinal $<|A|$. If $A$ is a domain, we are done; otherwise, there exists $i,j$ such that the projection on $A_i\times A_j$ is not a domain. Since $A_i$ and $A_j$ have the same characteristic, it follows that this projection is not cyclic, and hence, pulling back the cyclic subring, we obtain a proper unital subring index $<|A|$ in $A$.

Next, a reduced scalar ring $A$ is never Jónsson. Indeed, let $J$ be the set of prime numbers $p$ that are not invertible in $A$. For every $p$, let $P_p$ be the intersection of all prime ideals of $A$ containing $p$ (so $P_p=A$ for $p\notin J$). Then $A/P_p$ is $p$-reduced; if $|A/P_p|=|A|$ then $A/P_p$ is Jónsson and this contradicts the preceding paragraph. So $|A/P_p|<|A|$; hence the inverse image of its cyclic subring has index $<|A|$; since $A$ is Jónsson, this means that $A/P_p\simeq\mathbf{Z}/p\mathbf{Z}$ for every $p\in J$.
By the preceding paragraph, $A$ has at most one prime ideal $P_0$ such that $A/P_0$ has characteristic zero. By the case of domains, $|A/P_0|<|A|$. Hence, again by the argument of pulling back cyclic subrings, we see that $A/P_0$ is an infinite cyclic ring. Hence, if $P_0$ exists, it is contained in $P_p$ for every $p$, and in particular equals the intersection of all prime ideals, i.e., the nilradical, so $P_0=\{0\}$ since $A$ is reduced. Since $A$ is uncountable, we get a contradiction: $P_0$ does not exist. Hence the nilradical $\{0\}$ is equal to $\bigcap_{p\in J}P_j$. That is, the diagonal map $A\to\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}$ is injective.
Consider the composite map $A\to\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}\to (\prod_{p\in J}\mathbf{Z}/p\mathbf{Z})/(\bigoplus_p\mathbf{Z}/p\mathbf{Z})=B$; it has a countable kernel and hence an image of cardinal $|A|$. Then $B$ is a reduced scalar $\mathbf{Q}$-algebra, hence is 0-reduced. Since the image of $A$ in $B$ has cardinal $|A|$, we are done. (Alternative argument for these last few lines: the set of prime ideals of every scalar ring is compact, and it easily follows that if there exists prime ideals such that the quotient have unbounded characteristic, then there exists a prime ideal with quotient of zero characteristic.)

Now let us deal with $A$ arbitrary uncountable scalar ring; let $R$ be its nilradical. If $A/R$ has cardinal $|A|$, then $A$ is not Jónsson, by the reduced case, and if $A/R$ has cardinal $<|A|$ and is non-cyclic, then it has a proper unital subring of index $<|A|$. So we can suppose that $A/R$ is cyclic.
If $A/pA$ is has cardinal $|A|$ for some prime $p$, we can pass to $A/pA$. Otherwise $A/pA$ is has cardinal $<|A|$ for all $p$, and hence $A/nA$ has cardinal $<|A|$ for every $n\ge 1$; in particular, $A$ has characteristic zero, so no nonzero element of $\mathbf{Z}1_A$ is nilpotent: thus $A=R\oplus \mathbf{Z}1_A$. Since this is also true when $pA=0$, we henceforth suppose that $A=R\oplus \mathbf{Z}1_A$ ($A$ being of characteristic zero or prime).
Then observe that for every (non-unital) subalgebra $S$ of $R$, $S\oplus\mathbf{Z}1_A$ is a unital subalgebra of $R$.
If $R\neq 0$, there exists $x\in R$ with $x^2=0\neq x$. The kernel $K$ and image $I$ of the multiplication-by-$x$ map on $R$ are both ideals of $R$, and this multiplication induces a bijection $R/K\to I$. So either $I$ or $K$ has cardinal $|A|$. Hence both $K\oplus\mathbf{Z}1_A$ and $I\oplus\mathbf{Z}1_A$ are proper unital subalgebras of $A$, and at least one of them has cardinal $|A|$. So $R$ is not Jónsson.

Consequence: if $A$ is an associative unital ring, and is residually of cardinal $<|A|$ (that is, embeddable as unital subring of a product of unital rings of cardinal $<|A|$), then $A$ is not Jónsson.

(Note that Boolean algebras are residually finite, so this is a generalization.)

Indeed, the non-existence of any proper unital subring of countable index implies that every countable quotient of $A$ is cyclic; residual countability then implies that $A$ is commutative, and the previous case discards this.

Of course the argument says more, since it says that every associative unital ring $A$, which is residually of cardinal $<|A|$ (resp. residually countable, resp. residually finite), has a proper unital subalgebra of index $<|A|$ (resp. countable index, resp. finite index), except possibly in the case where $A$ embeds as a unital ring into the quotient of $\widehat{\mathbf{Z}}$ by some closed ideal.

Jonsson groups, rings and algebras.Irish Math. Soc. Bull.No. 36 (1996), 34–45. The author's name appears is spelled OREN KOLMAN on his homepage. $\endgroup$ – Ali Enayat Jul 12 '11 at 15:00