Morita equivalence of algebras certainly don't preserve commutativity: even if $A$ is commutative there are plenty of noncommutative algebras which are Morita equivalent with $A$---for example all algebras of the form $M_n(A)$ are good. What is the simplest example of an algebra which is not Morita equivalent to any commutative algebra?
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An algebra is Morita equivalent to a commutative algebra iff it's Morita equivalent to its center, since the center is Morita invariant. So any representative of a nontrivial class in the Brauer group of the underlying field $k$ is a counterexample: for example, when $k = \mathbb{R}$ we can take the quaternions $\mathbb{H}$.
answered Dec 14, 2014 at 21:52
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3$\begingroup$ I should mention that nothing preceding the example is needed to justify the example, only to suggest where to look for it: it's clear that $\mathbb{H}$, and more generally any noncommutative division ring, is not Morita equivalent to a commutative algebra because the endomorphism algebra of every $\mathbb{H}$-module is noncommutative. $\endgroup$ Commented Dec 15, 2014 at 6:48
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$\begingroup$ Thank you for your answer: I found that in fact: if $A$ and $B$ are Morita equivalent then their centers are isomorphic. Therefore it is enough to take a simple algebra $A$: if $A$ is Morita equivalent to commutative $B$ then $B=k$ (underlying field) and it is known that $A$ must be of the form $End_B(P)$ where $P$ is finitely generated $B$ module. In this case ($B=k$) this means that $A$ is a matrix algebra. $\endgroup$ Commented Dec 19, 2014 at 23:49
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1$\begingroup$ @truebaran: yes, that's what I meant by "the center is Morita invariant." $\endgroup$ Commented Dec 20, 2014 at 3:13