Let $X=S^2\vee S^2$ (wedge sum). The homology groups are $H_0(X,\mathbb{Z})= \mathbb{Z}$, $H_1(X,\mathbb{Z})= 0$, and $H_2(X,\mathbb{Z})= \mathbb{Z} \oplus\mathbb{Z}$.
I can see that $X$ is not homotopy equivalent to a closed manifold, since we would need a $\mathbb{Z}$ or $0$ in top homology (orientable, not orientable, resp) and torsion of $0$ or $\mathbb{Z_2}$ in the next lower homology (orientable, not orientable, resp), and we don't have that here.
So if $X$ is homotopy equivalent to some manifold, $M$, it must be one with boundary. The boundary would have to be composed of at least three components I believe... I'm thinking I could "fill out the two spheres to a 3 dimensional ball with the interiors of two disjoint open balls removed. This would be compact, 3-d manifold with boundary. However this looks like it isn't homotopy equivalent.
I also can see that by homotopy equivalence there are induced homomorphisms that are the identity on $H_*(X)$ and $H_*(M)$,but which pass through the other spaces homology. This gives that dim of $M$ must be atleast 3, but I'm guessing this allows me to say more...
For any finite complex $X$ that is embedded in $\mathbb{R}^n$ in a piecewise smooth manner, $X$ has a regular neighborhood which is a compact $n$-manifold with boundary that deformation retracts to $X$.
In the case of the standard embedding $S^2 \vee S^2 \subset \mathbb{R}^3$, that regular neighborhood is as described by @DanielRust, and consists of large compact 3-ball from which two small open 3-balls have been removed, whose closures are disjoint subsets of the interior of the large ball.