Simply-connected $\mathbb{Z}_p$-homology spheres?

Let $X$ be a $\mathbb{Z}$-homology $n$-sphere, i.e., a closed manifold with $\mathbb{Z}$-homology groups of the standard $n$-dimensional sphere. If $X$ is simply-connected, it is not difficult to see (and not difficult to look up on the internet) that such a space is in fact homotopy equivalent (hence homeomorphic, by the Poincare conjecture) to the $n$-sphere.

Now assume that $X$ is a $\mathbb{Z}_p$-homology sphere for some prime $p$. Is it true that if $X$ is simply-connected, then $X$ is homotopy equivalent to the $n$-sphere?

EDIT: OOPS, sorry! As pointed out in the comments, the first sentence didn't make any sense before. Added simply-connecteness as an assumption.

As pointed out by Qiaochu Yuan, there are counterexamples to your claim in dimension five. The simplest such counterexample is the Wu manifold, constructed below, which is a simply connected $\mathbb{Z}_p$-homology sphere for every odd prime $p$ (in fact, every odd integer $p$) but is not homeomorphic to a sphere (in particular, it is not a $\mathbb{Z}_2$-homology sphere).

As $SO(3)$ is a closed Lie subgroup of $SU(3)$, the set of left cosets of $SO(3)$, namely $M := SU(3)/SO(3)$, is a connected, closed manifold and the projection map $\pi : SU(3) \to M$ is a principal $SO(3)$-bundle. Note that $\dim M = \dim SU(3) - \dim SO(3) = 8 - 3 = 5$. From the long exact sequence in homotopy, we see that

$$\dots \to \pi_2(SU(3)) \to \pi_2(M) \to \pi_1(SO(3)) \to \pi_1(SU(3)) \to \pi_1(M) \to \pi_0(SO(3)) \to \dots$$

As $SU(3)$ is simply connected and $SO(3)$ is connected, it follows that $M$ is simply connected. Every Lie group has trivial second homotopy group, so $\pi_2(M) \cong \pi_1(SO(3)) \cong \mathbb{Z}_2$. By the Hurewicz Theorem, $H_1(M; \mathbb{Z}) = 0$ and $H_2(M; \mathbb{Z}) \cong \mathbb{Z}_2$.

Note that $M$ is orientable (because $M$ is simply connected), so by Poincaré duality $H_4(M; \mathbb{Z}) \cong H^1(M; \mathbb{Z}) = \operatorname{Hom}(\pi_1(M), \mathbb{Z}) = 0$. By the Universal Coefficient Theorem, $H^2(M; \mathbb{Z}) \cong \operatorname{Hom}(H_2(M; \mathbb{Z}_2), \mathbb{Z})\oplus\operatorname{Ext}(H_1(M; \mathbb{Z}_2), \mathbb{Z}) \cong \operatorname{Hom}(\mathbb{Z}_2, \mathbb{Z})\oplus\operatorname{Ext}(0, \mathbb{Z}) = 0$ so, again by Poincaré duality, $H_3(M; \mathbb{Z}) \cong H^2(M; \mathbb{Z}) = 0$.

Therefore $M$ is a simply connected, five-dimensional manifold with integral homology groups

$$\mathbb{Z}, 0, \mathbb{Z}_2, 0, 0, \mathbb{Z}.$$

Let $p$ be an odd integer (e.g. a prime other than $2$). As $\mathbb{Z}_2\otimes\mathbb{Z}_p = 0$ and $\operatorname{Tor}(\mathbb{Z}_2, \mathbb{Z}_p) = 0$, the Universal Coefficient Theorem for homology shows that $M$ has $\mathbb{Z}_p$ homology groups

$$\mathbb{Z}_p, 0, 0, 0, 0, \mathbb{Z}_p.$$

So $M$ is a simply connected $\mathbb{Z}_p$-homology sphere which is not homeomorphic to a sphere.

Edit #2: There is a $5$-dimensional counterexample, but I don't know how to produce it explicitly.

According to Smale's classification of closed, oriented, smooth, simply connected, spin $5$-manifolds, there is a simply connected closed $5$-manifold $X$ with $H_2(X) \cong \mathbb{Z}_p \oplus \mathbb{Z}_p$. By the same universal coefficient and Poincaré duality arguments as below, such a manifold $X$ is a $\mathbb{Z}_q$-homology sphere for any prime $q \neq p$, but not a $\mathbb{Z}$-homology sphere.

Universal coefficients also shows that a simply connected closed $n$-manifold which is a $\mathbb{Z}_p$-homology sphere for all primes $p$ is a $\mathbb{Z}$-homology sphere.

Edit: This is not an answer. I just want to record the observation that the minimal dimension of a counterexample is $5$.

First let me record that a simply connected closed manifold cannot have a nontrivial orientation double cover, and hence is orientable.

The cases $n = 0, 1$ are degenerate. When $n = 2$ the $2$-sphere is already the only simply connected closed surface by the classification of surfaces.

When $n = 3$, let $X$ be a simply connected closed $3$-manifold. Since it's orientable, $H_3(X) \cong \mathbb{Z}$ and $H_2(X) \cong H^1(X) \cong 0$ by Poincaré duality, and hence $X$ is already a $\mathbb{Z}$-homology sphere.

When $n = 4$, let $X$ be a simply connected closed $4$-manifold. Since it's orientable, $H_4(X) \cong \mathbb{Z}$. By universal coefficients, $H^2(X)$ is torsion-free, so by Poincaré duality, $H_2(X) \cong H^2(X)$ is also torsion-free. And by Poincaré duality again, $H_3(X) \cong H^1(X) \cong 0$. It follows that

$$H_2(X, \mathbb{Z}_p) \cong H_2(X) \otimes \mathbb{Z}_p$$

vanishes iff $H_2(X)$ vanishes. Hence if $X$ is a simply connected closed $4$-manifold which is also a $\mathbb{Z}_p$-homology sphere for any $p$, then $X$ is a $\mathbb{Z}$-homology sphere.