I'm not 100% sure this is correct, but here goes...
I first claim that there is a simply connected space $X$ whose reduced homology vanishes with $\mathbb{Z}[\frac{1}{2}]$-coefficients, and map $X \to BSO$ inducing a map of Thom spectra $MX \to MSO$ whose image on homotopy groups is precisely the 2-torsion.
Proof of claim. If $X$ has torsion homology then so does $MX$ and hence torsion homotopy. Thus for any such $X$ we can never hit more than the torsion subgroup of $\pi_*(MSO)$, 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 $$\Omega^2 S^3 \to \Omega^2 \Sigma^2 \mathbb{R}P^\infty \to BO.$$ Now let $Y = \Omega^2 \Sigma^2 \mathbb{R}P^\infty$, take 1-connected cover and write $X = \tau_{\geq 2} Y$, Thomify, 2-localize, and use the Hopkins--Mahowald theorem to get maps of $E_2$ ring spectra $$H \mathbb{Z} _{(2)} \to MX_{(2)} \to MSO_{(2)}.$$ Thom isomorphism identifies $H_*(MX;\mathbb{F}_2) \to H_*(MSO;\mathbb{F}_2)$ with the map induced by $X = \tau_{\geq 2}Y \to \tau_{\geq 2} BO = BSO$, which is still surjective on mod 2 homology (since both $Y$ and $BO$ split as $\mathbb{R} P^\infty$ times their 1-connected covers). 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. Both modules are free on the image of the Hurewicz homomorphism from mod 2 homotopy, which is injective. Hence we may apply $\mathbb{F}_2 \otimes_{\mathcal{A}_2^\vee} -$ to see that the induced map on mod 2 homotopy is 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 later purposes 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 $H_n(M;\mathbb{Q})$. I haven't thought through the case where $d = 2n+1$, but assume it's similar.
If that all works out, the conclusion should be that all the torsion elements are represented by rational homology spheres.