Well here's my shot: (skip to the punchline at the bottom if you want)

Take an associative algebra A and a k-local ring R (the formal power series over k, or the infinitesimal ring will do nicely).

The algebra A is naturally a homotopy algebra and so may be given by a degree -1 square-zero coderivative on the free coassociative coalgebra on A[1]. We write this coalgebra BA, the bar resolution. Note that in homotopy theory it often makes life easier if we forget any unit elements; BA is non-unital.

An A-infty R-deformation of A is now a square-zero coderivative on the coalgebra BA⊗R, such that the "obvious" diagram commutes (I could post this as an image when I'm permitted). The condition could alternatively by phrased as the following:
"such that it extends the original coderivative on BA".

So far this has all been definitions, my answer to your question comes next:
Consider now the cobar functor applied to the morphism BA⊗R→BA,

Ω(BA⊗R) ≅ (ΩBA)⊗R → ΩBA.

This is a proper algebra deformation, nothing infinity about it! Except... ΩBA is homotopy equivalent to A.

The short and snappy answer:

## Infinity deformations are homotopy invariant, classical algebra deformations are not.

Edit: I should have added, if you would like me to expand on anything, I'm more than willing.