@Tomek: Thank you for pointing out that reference, I was not aware of it. It deals exaclty with the topic I asked about.

As Taka pointed out, among the simple, separable, non-type I C*-algebras, all the nuclear ones are isomorphic as Banach spaces (even much more is true) but beyond the nuclear case this is no longer the case. This certainly answers my question, thanks.

@Caleb: Thanks for pointing out the result of Kirchberg. This clarifies the picture a lot. The question now seems to be whether the Banach space structure can detect nuclearity of the C*-algebra.

@Qiaochu: Thanks for clarifying. Here, two Banach spaces E and F are called isomorphic if there exists a bounded linear map from E to F that is bijective. (The inverse map will automatically be bounded, too). I do not want to assume isometric isomorphism.

@Yves: Thank you for the examples. I also agree with you (and Ramiro) that the Suslin condition is too strong and should be relaxed to what you suggest.

Thank you Anton, I agree that the ($\epsilon$,$\delta$)-covering property that you mention is preserved by uniform equivalence. I just remark that it reminds me of the box-counting dimension of metric spaces, and this dimension is preserved by Lipschitz equivalence, but I think not by uniform equivalence (of course, you didn't claim that). Do you think that the ($\epsilon$,$\delta$)-covering property (together with finite uniform covering dimenion) will be sufficient for uniform embedding into Euclidean space? Is this covering property totally independent of the uniform covering dimension?

Emil, I agree. Every closed subset of $\mathbb{R}^n$ is complete, and uniform equivalence of metric spaces preserves completeness. Thus, when asking for spaces that are uniformly equivalent to closed subsets of Euclidean space, one has to add completeness to the list of neccessary conditions. But it is maybe more natural to drop the assumption that $Y\subset\mathbb{R}^n$ be closed.