Complementing the other answers in this thread: while the cohomology ring of a simply connected space does not determine the cohomology of the loop space, the rational cohomology viewed as an $A_\infty$-algebra does.
Namely, the cohomology of any $A_\infty$-algebra $A$ over $\mathbf{Q}$ (in particular, of any differential graded algebra) carries an $A_\infty$-structure such that there is an $A_\infty$ map $H^{*}(A)\to A$ inducing the identity in cohomology; this $A_\infty$ structure is unique up to a non-unique isomorphism. See e.g. Keller, Introduction to A-infinity algebras and modules, 3.3 and references therein. By taking $A$ to be the rational singular cochains of a topological space $X$ we get an $A_\infty$-structure on $H^{*}(X,\mathbf{Q})$.
To each $A_\infty$ algebra $B$ H$ there corresponds a bar construction, which is a free differential coalgebra on $B$ H$ shifted by 1 to the left (see e.g. 3.6 of Keller's paper entioned above). It is an old result of Kadeishvili (see MR0580645) the that if $B$ H$ is the cohomology of a simply-connected space $X$ with the above $A_\infty$-structure, then the cohomology of the bar construction is the cohomology of the bar construction is the cohomology of $\Omega (X)$.
This also explains why we should expect a negative answer to the question as it is stated: all components of the $A_\infty$ structure on the cohomology participate in the bar construction, and not just the product.
