Ryan Budney's comment pretty much killed the question, but anyway...

Let $X_{m,n} = S^{2m}\times S^{2n+1}$, with $m\le n$ (strictly) positive integers.

**Lemma**: $X=X_{m,n}$ is parallelisable.

*Proof*: this follows from playing around with vector bundles, the key facts being that $TX = \pi^*(TS^{2m}) \oplus \pi^*(TS^{2n+1})$ and that trivial bundles are natural.
More precisely, the second factor of $X$ has Euler number 0, so one can split off a trivial 1-bundle from its tangent bundle, pull it back through the projection and see it as the pull-back of a trivial bundle over the first factor. So $TX = \pi^*(TS^{2m}\oplus \varepsilon) \oplus V$, and the first summand is trivial. Now one can split off a trivial bundle of rank *2* from the first factor and use it to trivialise the second factor.

**Lemma**: $X$ is not an $H$-space.

*Proof*: the cohomology ring (say with $\mathbb{F}_2$ coefficients) of an $H$-space is a finite-dimensional commutative Hopf algebra, therefore it's generated in odd dimensions. But, in our case, $H^*(X)$ has $H^{2m}(X)$ as the first nontrivial group.

This gives a nice family of simply connected counterexamples to your question, the smallest of which is $S^2\times S^3$. Notice how, actually, there is no simply connected counterexample in dimension one, two, three (thanks to Perelman) and four (*e.g.* because the Euler number of a simply connected 4-manifold is strictly positive).