Fix a prime $p$ and consider everything mod $p$. Steenrod operations arise somehow from the loss of information passing from the singular complex of a space to its cohomology ring. Are they exactly this gap, i.e. can I get the singular complex back from the cohomology ring of a space and its structure as a module over the Steenrod algebra?

No. For instance, Massey products on the cohomology are extra information that neither the ring structure nor the Steenrod operations see. The complement of the Borromean rings, for example, and the complement of three unlinked circles in $R^3$ have the same cohomology ring and Steenrod operations, but cannot be chain equivalent because of nonzero Massey products in the former. 


This is answering a slightly different question, but here goes: If the question is "does knowing the singular (co)chains up to quasiisomorphism determine the space up to weak equivalence?" then of course the answer is no. To some extent the answer can be turned into a yes by altering the question, considering the (co)multiplication on (co)chains as part of the structure that a quasiisomorphism must preserve. Mike Mandell has made this idea precise (using operads) and proved powerful theorems along these lines, but I'm not an expert. Of course, the operad structure on chains of course determines the cup product, the Steenrod operations, and more (e.g. differentials in Adams spectral sequence). EDIT: And Massey products. 


It depends of what structure you want to consider on the complex of singular cochains. If you want to look at it just as a complex, then the cohomology groups are enough. If you want it as a differential graded algebra, then you would need the cohomology groups with an Ainfinity algebra structure, etc. Steenrod operations are something which come from the Einfinity structure on cochains, but they are weaker. 


I think the answer is no, due to this answer to this question. There are Moore spaces with the same cohomology rings and module structure over the mod $p$ Steenrod algebra, which nevertheless have different homology groups and so their singular complexes cannot be weakly equivalent. 

