Let $X$ be the sort of topological space for which it makes sense to talk about the intersection homology. Fix a perversity $p$, or just take $p= 1/2$ if you like.
Is there some naturally defined $X'$ such that ${}^p IH_* (X) = H_* (X')$ ?
Let $X$ be the sort of topological space for which it makes sense to talk about the intersection homology. Fix a perversity $p$, or just take $p= 1/2$ if you like.



I'm not sure this is the kind of answer you want, but if $X$ has a small resolution $f:X' \rightarrow X$ (so $X'$ is a manifold and the dimension of fibers is sufficiently small), then there is an induced isomorphism $IH_{\ast}(X) = IH_{\ast}(X')$, and because $X'$ is smooth, the latter group is $H_{\ast}(X')$. More precisely, a proper (birational) map is small if the set $${x \in X  \dim f^{1}(x) \geq r }$$ has codimension more than $2r$; such maps induce isomorphisms on IH. Of course, there is nothing natural about $X'$, nor do small resolutions necessarily exist (see the comment below by Mike Skirvin for an easy example). Hopefully someone more knowledgeable about IH will have something to say. 


The short answer is "no". But you might be interested in some work Markus Banagl is doing along these lines. He has a functor that assigns to a space X an "intersection space" $I^{\bar p}X$. Then rather than studying $I^{\bar p}H_\ast(X)$, he looks at $H_\ast(I^{\bar p}X)$. However, this is not generally the same thing as $I^{\bar p}H_\ast(X)$. 


Such a space $X'$ cannot possibly depend functorially on $X$: suppose that there is a functor $F$ from the category of topological spaces (that have reasonable stratifications) and continuous maps to itself such that IH(X)=H(F(X)). Then IH would be functorial, but considering the example $$(S^1 \hookrightarrow X \rightarrow S^1)=id_{S^1}$$ where $X$ is the pinched torus and $S^1$ is the noncollapsed circle (see pages 55 and 56 of KirwanWoolf "Introduction to Intersection Homology") would imply that the intersection homology of $S^1$ (=ordinary homology of $S^1$) imbeds in the intersection homology of $X$. But the first IH group of $X$ (middle perversity) is $0$. 

