I don't have a complete answer, but here are a few remarks about this that you may find interesting:

You didn't specify exactly what you meant by 'complex hyperbolic space' $\mathbb{CH}^n$ and 'hyperbolic space' $\mathbb{H}^n$, in the sense that you didn't specify the sectional curvatures of the two spaces.  I will take the meaning of $\mathbb{H}^n$ to be the simply-connected complete Riemannian manifold of constant sectional curvature $-1$ and $\mathbb{CH}^n$ to be the simply-connected, complete Hermitian symmetric space of constant holomorphic sectional curvature that contains  $\mathbb{H}^n$ as a totally geodesic real submanifold.  (Note that the  sectional curvature of tangent $2$-planes in $\mathbb{CH}^n$ that are complex lines is $-4$, not $-1$.  The sectional curvature function of $\mathbb{CH}^n$ takes values in the interval $[-4,-1]$.  By convention, the sectional curvature of $\mathbb{CH}^1$ is $-4$ (so that the natural complex linear embedding $\mathbb{CH}^1\subset\mathbb{CH}^2$ will be an isometry), 
so $\mathbb{CH}^1$ is not isometric to $\mathbb{H}^2$, but to $\mathbb{H}^2$ with its metric divided by $4$.)  

Now, the obvious thing to try to construct a Riemannian submersion from $\mathbb{CH}^n$ to $\mathbb{H}^n$ doesn't work:  The 'nearest point' projection from $\mathbb{CH}^n$ to the submanifold $\mathbb{H}^n\subset \mathbb{CH}^n$ (which *is* a smooth submersion) is *not* a Riemannian submersion.

The next obvious thing to try (especially since you want a 'canonical' Riemannian submersion) is to look for one that is as homogeneous as possible.  Whether a 'homogeneous' example exists for all $n$ or not, I don't know, but a little calculation shows that such a Riemannian submersion *does* exist (and essentially uniquely) when $n=2$, the first nontrivial case.  More precisely, one has the following result:

**Fact:**  (1) There exists a subgroup $G\subset\mathrm{Isom}(\mathbb{CH}^2)$ that acts simply transitively on $\mathbb{CH}^2$, a homomorphism $\rho:G\to\mathrm{Isom}(\mathbb{H}^2)$ such that $\rho(G)$ acts transitively on $\mathbb{H}^2$, and a Riemannian submersion $\pi:\mathbb{CH}^2\to \mathbb{H}^2$ such that $\pi\bigl(g(m)\bigr) = \rho(g)\bigl(\pi(m)\bigr)$ for all $m\in\mathbb{CH}^2$.  (2) Moreover, this homogeneous Riemannian submersion $\pi$ is unique up to composition with isometries in $\mathbb{CH}^2$ and $\mathbb{H}^2$: If $\widehat\pi:\mathbb{CH}^2\to\mathbb{H}^2$ is a Riemannian submersion with the property that the group $G\subset \mathrm{Isom}(\mathbb{CH}^2)\times \mathrm{Isom}(\mathbb{H}^2)$ consisting of the pairs $(g,h)$ such that $\widehat\pi\bigl(g(m)\bigr) = h\bigl(\widehat\pi(m)\bigr)$ for all $m\in\mathbb{CH}^2$ acts transitively on $\mathbb{CH}^2$, then $\widehat\pi = h\circ \pi\circ g$ for some $(g,h)\in \mathrm{Isom}(\mathbb{CH}^2)\times \mathrm{Isom}(\mathbb{H}^2)$.


Meanwhile, there exist many Riemannian submersions $\pi:\mathbb{CH}^2\to\mathbb{H}^2$ that are not homogeneous.  For example, there exist examples whose commuting isometry group $G\subset \mathrm{Isom}(\mathbb{CH}^2)\times \mathrm{Isom}(\mathbb{H}^2)$ (as defined above) acts in cohomogeneity $1$ or $2$ on $\mathbb{CH}^2$.

If you are interested, I can supply details.