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Better argument
Denis Nardin
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This is an assemblage of known results, I'll try to put a reference for all of them.

  • By a classical theorem of Serre all stable homotopy groups are finite in positive degree. In particular we have $\mathbb{S}_\mathbb{Q}=H\mathbb{Q}$.

  • Rationalization is a smashing localization, so that $\pi_*(\mathbb{S}_\mathbb{Q}\otimes E)\cong \pi_*E \otimes_{\mathbb{Z}}\mathbb{Q}$. This follows immediately from the fact that we can write $$E_\mathbb{Q}=\mathrm{colim}\left(E\xrightarrow{2}E\xrightarrow{3}E\xrightarrow{4}E\xrightarrow{5}\cdots\right)$$ and that tensor (smash) products and homotopy groups commute with filtered colimits of spectra.

  • By Schwede-Shipley Morita theory, the category of modules over $H\mathbb{Q}$ is equivalent to the derived category of $\mathbb{Q}$. (Theorem 5.1.6 in Stable model categories are categories of modules) Moreover under this equivalence homotopy groups correspond to homology groups (because $H\mathbb{Q}$ is sent to $\mathbb{Q}$ in degree 0).

  • In the derived category $D(\mathbb{Q})$ for every object $M$ there is an equivalence $M\cong \bigoplus_{n\in\mathbb{Z}} (H_nM)[n]$. This is an easy exercise, using the fact that in $\mathbb{Q}$-vector spaces every short exact sequence splits.

In particular for every spectrum $E$, the spectrum $E\otimes H\mathbb{Q}$ has homotopy groups $\pi_*(E\otimes H\mathbb{Q})\cong \pi_*E\otimes_{\mathbb{Z}}\mathbb{Q}$ and so there is an equivalence of $H\mathbb{Q}$-modules (in particular of spectra)

$$ E\otimes H\mathbb{Q}\cong \bigoplus_{n\in\mathbb{Z}} \Sigma^nH(\pi_nE\otimes_{\mathbb{Z}}\mathbb{Q})\cong\prod_{n\in\mathbb{Z}} \Sigma^nH(\pi_nE\otimes_{\mathbb{Z}}\mathbb{Q})$$ (it's not necessary, but it is very convenient to observe that in this case for degree reasons the coproduct and the product coincide)

So, to conclude, we have that $$[bstring, X\otimes H\mathbb{Q}]\cong \left[ bstring, \prod_{n\in\mathbb{Z}}\Sigma^nH(\pi_nX\otimes_{\mathbb{Z}}\mathbb{Q})\right]\cong \prod_{n\in\mathbb{Z}} H^n(bstring;\pi_nX\otimes_{\mathbb{Z}}\mathbb{Q})\cong \prod_{n\in\mathbb{Z}} \pi_nX\otimes_{\mathbb{Z}} H^n(bstring;\mathbb{Q})\,.$$

Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103