(Building on Goldstern's comment:)  If fields are ok, (and if you allow _infinite_ algebras -- see comment by Mariano Suárez-Alvarez below) then the distributivity certainly does not hold.

Take e.g. a finite degree extension $F$ of $\mathbb{Q}$ with Galois group $G$.  Subalgebras of $F$ are subfields, by undergraduate field theory, so the subalgebra lattice over $\mathbb{Q}$ is the field extension lattice of $F:\mathbb{Q}$.  (I'm assuming that you're taking all your algebras over a fixed field, here $\mathbb{Q}$.)  By the Galois correspondence, the field extension lattice is anti-isomorphic to the subgroup lattice of the Galois group.


And subgroup lattices certainly need not be distributive.  Indeed, by a theorem of Ore, the subgroup lattice $L(G)$ of a finite group $G$ is distributive iff $G$ is cyclic.