I think that the answer is no, but possibly for dumb reasons. There is a model for $QX$ (when $X$ is connected):

$$QX \simeq \coprod_n E\Sigma_n \times_{\Sigma_n} X^{\times n} / \sim$$

where $\Sigma_n$ is the symmetric group, and so $E\Sigma_n$ is taken to be the ordered configuration space of $n$ points in $\mathbb{R}^\infty$. We think of this as the space of points in $\mathbb{R}^\infty$ labelled by points of $X$. The relation $\sim$ drops points when their label is the basepoint.

Snaith has shown that this space stably splits along the filtration by the number $n$. When $X = S^k$ (with $k>0$), the split summands are identifiable as the Thom space $B\Sigma_n^{k\gamma_n}$, where $\gamma_n = \mathbb{R}^n$ is the permutation representation of $\Sigma_n$, and $k\gamma_n$ is its $k$-fold sum. Its homology is then a shift by $nk$ of the homology of $\Sigma_n$ with coefficients in the $k$th tensor power of the sign representation:

$$\tilde{H}_{*+nk}(B\Sigma_n^{k\gamma_n}) = H_*(\Sigma_n, sign^{\otimes k})$$

since the action of $\Sigma_n$ on the top dimensional homology of $\gamma_n \cup \{ \infty\}$ is by the sign representation. In particular, if $k$ is even, this is the trivial representation.

So take $X = S^2$; it has a homology exponent $0$, since $1 = p^0$ annihilates all of the $p$-torsion in $H_*(X)$. However, the homology of $QX$ contains $H_*(S_n, \mathbb{Z})$ for all $n$. There is an argument that I learned in Remark 1.11 of this nice paper by Pakianathan which shows that the homology $p$-exponent of the Sylow $p$ subgroup $S(p^n) \leq \Sigma_{p^n}$ is $p^n$. Now take an element in $H_*(S(p^n), \mathbb{Z})$ realizing that exponent, and transfer it up to an element $x_n \in H_*(\Sigma_{p^n})$ to get a family of elements $\{x_1, x_2, \dots\}$ of unbounded homology $p$-exponent in $H_*(QS^2)$.

Now, if you restrict your attention to connected spaces $X$ with entirely torsion homology, I think that if $X$ has a $p$-exponent, then so does $QX$. The reason is that we can again use the Snaith splitting to get

$$H_*(QX) = \bigoplus_n H_*(E\Sigma_{n+} \wedge_{\Sigma_n} (X^{\wedge n}))$$

and you can compute $H_*(E\Sigma_{n+} \wedge_{\Sigma_n} (X^{\wedge n}))$ as the target of a spectral sequence whose $E_2$ term is $H_*(\Sigma_{n}, \tilde{H}_*(X^{\wedge n}))$. If the homology $p$-exponent of $X$ is $m$, then $p^m$ kills the module $\tilde{H}_*(X^{\wedge n})$ (using the Künneth theorem), and hence this group homology. There might be some extension problems in the SS that I'm ignoring here, but if they can be resolved, then $p^m$ must kill $H_*(QX)$.