If the following is correct, it shows that *any* torsion class in $\Omega_*^{SO}$ can be represented by a rational homology sphere.

I first claim that there is a simply connected space $X$ such that $\widetilde{H}_*(X;\mathbb{Z}[\frac12]) = 0$, and map $X \to BSO$ such that on homotopy groups of the induced map of Thom spectra, the image of $\pi_d(MX) \to \pi_d(MSO) = \Omega_d^{SO}$ is precisely the torsion, for $d > 0$.

Proof of claim.  Finiteness of the stable homotopy groups of spheres implies that $\pi_d(MX)$ is a torsion group for $d > 0$ for any such $X$.  Therefore we can never hit more than the torsion subgroup of $\pi_d(MSO)$, all of which is known to be 2-torsion.  The difficult part is proving that this may indeed all be hit.

The non-trivial based map $S^1 \to BO$ factors through $\mathbb{R} P^\infty \to BO$, whose image in mod 2 homology generates the Pontryagin ring $H_*(BO;\mathbb{F}_2)$.  We can freely extend to double loop maps
$$\Omega^2 S^3 \to \Omega^2 \Sigma^2 \mathbb{R}P^\infty \to BO$$
where the second map then induces a surjection on mod 2 homology.  Both $\Omega^2 \Sigma^2 \mathbb{R}P^\infty$ and $BO$ split as $\mathbb{R} P^\infty$ times their 1-connected cover, so the induced map of 1-connected covers $\tau_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty) \to \tau_{\geq 2}(BO) = BSO$ also induces a surjection on mod 2 homology.

Now let $X = \tau_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty)$ with the map to $BSO$ constructed above.  Take 1-connected covers of the double loop maps above, Thomify, 2-localize, and use the Hopkins-Mahowald theorem to get maps of $E_2$ ring spectra 
$$H \mathbb{Z} _{(2)} \to MX_{(2)} \to MSO_{(2)}.$$
(See e.g. section 3 of [this paper][1].)
We can view $MX_{(2)} \to MSO_{(2)}$ as a map of $H\mathbb{Z}_{(2)}$-modules, and the induced map on mod 2 homology inherits the structure of a module map over the mod 2 dual Steenrod algebra $\mathcal{A}^\vee$.  Both modules are free: in fact the mod 2 Hurewicz homomorphism $\pi_*(MX;\mathbb{Z}/2\mathbb{Z}) \to H_*(MX;\mathbb{Z}/2\mathbb{Z})$ induces an isomorphism
$$\mathcal{A}^\vee \otimes \pi_*(MX;\mathbb{Z}/2\mathbb{Z}) \to H_*(MX;\mathbb{Z}/2\mathbb{Z}),$$
and similarly for $MSO$.  Therefore the map $\pi_*(MX;\mathbb{Z}/2\mathbb{Z}) \to \pi_*(MSO;\mathbb{Z}/2\mathbb{Z})$ may be identified with the map obtained by applying $\mathbb{F}_2 \otimes_{\mathcal{A}^\vee} -$ to the map on homology, which is surjective by Thom isomorphism, showing that the induced map on mod 2 homotopy is also surjective.  Now any 2-torsion class $x \in \pi_d(MSO)$ comes from $\pi_{d+1}(MSO;\mathbb{Z}/2\mathbb{Z})$, hence from $\pi_{d+1}(MX;\mathbb{Z}/2\mathbb{Z})$ and in particular from $\pi_d(MX)$. $\Box$

We have shown that any torsion class $[M] \in \Omega_d^{SO}$ admits a representative where the structure map $M \to BSO$ lifts to $f: M \to X$.  By the usual arguments (that is, doing surgery on representatives of $\pi_*(X,M)$) there is no obstruction to making this map be $n$-connected for $d = 2n$ or $d = 2n+1$.  For the purposes of a later induction argument, let me weaken the conclusion to $f$ being $(n-1)$-connected and rationally $n$-connected.  To get to a rational homology sphere, it remains to make $H_n(M;\mathbb{Q}) = 0$.   If this is not already the case we may, in the case $d = 2n$ where the signature must vanish when $n$ is even, choose a non-zero Lagrangian element $x \in H_n(M;\mathbb{Q})$.   After possibly multiplying by a positive integer, the rational Hurewicz theorem implies that this element may be lifted to $\pi_{n+1}(X,M)$ and be represented by an embedded $S^n \to M$ with trivial normal bundle.  Doing surgery on that framed embedding gives a new manifold $M'$ with $H_n(M';\mathbb{Q})$ smaller than $H_n(M;\mathbb{Q})$.  Unfortunately the surgery process may create some undesired torsion in $H_{n-1}(M')$ if the sphere we do surgery on is divisible in $H_n(M)/\mathrm{torsion}$ (the new torsion is generated by a meridian $(n-1)$-sphere to the sphere we do surgery on; compare the diagram on page 517 of Kervaire-Milnor) but $M' \to X$ will remain rationally $n$-connected.  Repeat this until $H_n(M;\mathbb{Q}) = 0$.

The case $d = 2n+1$ is similar.  If $H_n(M;\mathbb{Q}) \neq 0$ we may choose a non-zero element in that group, lift a multiple to $\pi_{n+1}(X,M)$, and represent by an embedded sphere with trivialized normal bundle.  Doing surgery on this framed embedding will make $H_n(M;\mathbb{Q})$ smaller, although it may create new torsion in $H_n(M)$ if the sphere we do surgery on is divisible in $H_n(M)/\mathrm{torsion}$ (compare the diagram on page 515 of Kervaire-Milnor).

  [1]: https://arxiv.org/abs/2004.06889