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.) 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 scaling 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 element we do surgery on is divisible (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. Continuing in this fashion should lead to $H_n(M;\mathbb{Q}) = 0$. I haven't thought through the case where $d = 2n+1$, but assume it's similar.