First of all, your question should be more carefully formulated, since it is not even possible to always find a non-trivial sub*system* (not sub*space*) of $\mathcal H_A$, see also my comment here https://quantumcomputing.stackexchange.com/questions/14049/can-a-single-qutrit-in-superposition-be-considered-entangled/14052#14052

Thus, let us assume that $\mathcal H_A$ does not have prime dimension, $d_A=\dim\mathcal H_A$. The question should then be the following: 
Given a Hilbert space $\mathcal H_B$ where $d_B=\dim\mathcal H_B$ divides $d_A$, and a density matrices $\rho\in\mathcal H_A$ and $\sigma\in\mathcal H_B$, is there a choice of tensor product $\mathcal H_A \simeq \mathcal H_B\otimes\mathcal H_C$, such that $\mathrm{tr}_C(\rho)=\sigma$.

I'm not so sure why you would call this a "purification", though.

This can sometimes be true. For example, take for simplicity $\mathcal H_B=\mathcal H_C=\mathbb C^d\otimes\mathbb C^d$ and define $\mathcal H_A := \mathcal H_B\otimes \mathcal H_C$.
Let $\phi^+$ be the standard maximally entangled state with respect to the bipartition of $\mathcal H_C=\mathcal H_B$. 
Consider the product state $\rho=|\phi^+\rangle\langle\phi^+|\otimes|\phi^+\rangle\langle\phi^+|$, and $\sigma = \mathbb{I}/d^2$. Then there is always a different bipartition $A=B'|C'$ of $A$, such that $\mathrm{tr}_{C'}(\rho) = \sigma$. You can explicitly verify this by writing down the state and swapping the second tensor factors of $B$ and $C$.

**Update:**
However, in general, this is false since the partial trace make the state $\rho$ "more mixed". Thus if the von-Neumann entropy of $\sigma$ is lower than the one of $\rho$, this should not be possible (hand-wavely spoken).

Here's a counter-example. Take $\rho = \mathbb{I}/d_A$. Then, for any choice of bipartition, $\mathrm{tr}_C(\rho) = \mathbb{I}/d_B$. Take $\sigma$ to be any other state, e.g. a pure one.