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. 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.