Let $\cal C$ be an $A_\infty$-category and $A$ its total algebra (elements in $A$ are formal linear combinations of arbitrary morphisms in $\cal C$ and multiplications of arrows which can't be concatenated are defined to be zero).

If $\cal C$ is formal, it is obvious that $A$ is formal, too. Does the converse hold?

The reason I think this could hold is as follows:

Let $(\bar A,\mu)$ be a graded algebra. Then one can consider the groupoid $M(A)$ of all $A_\infty$-structures which extend $(\bar A,\mu)$. More precisely objects are $A_\infty$-structures on $\bar{A}$ of the form $(0,\mu,\mu_3, \mu_4 \ldots)$ and morphisms are $A_\infty$-morphisms of the form $(id_{\bar A}, \psi_2,\psi_3,\ldots)$. This space has a distinguished point namely $(\bar A,\mu)$ and if I understand the experts correctly the tangent space at this point is the Hochschild Complex $CH(\bar{A},\bar{A})$. Also there is is an action of the multiplicative group by rescaling the higher multiplications, which contracts this space to the special point. An analogue holds if we replace the graded algebra by a graded category.

Lets apply this to the question: We can assume that $\cal C=(\cal{C},0,\mu,\mu_3,\ldots)$ has vanishing differential and consider the underlying graded category $\bar{\cal C}=(\cal C, \mu)$ and similar for the total algebra.

Now there is a morphism $M(\bar{\cal C}) \rightarrow M( \bar{A})$ which preserves the distinguished points and is equivariant for the contracting actions. Furthermore it induces a quasi-isomorphism between the tangent spaces (i.e. the hochschild complexes) and I would expect such a morphism to be an equivalence.

As a corollary an $A_\infty$-structure on $\bar{\cal C}$ would be formal iff the corresponding structure on $\bar A$ is formal.