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 $\mathbb{CH}^2$.
Details: Let $\mathrm{SU}(1,2)\subset\mathrm{GL}(3,\mathbb{C})$ be the connected subgroup such that its canonical left-invariant form has the expression $$ \gamma = g^{-1}\mathrm{d}g = \pmatrix{ -(\phi_{1\bar1}{+}\phi_{2\bar2}) & \overline{\omega_1} & \overline{\omega_2} \\ \omega_1 & \phi_{1\bar1} & \phi_{1\bar2}\\ \omega_2 & \phi_{2\bar1} &\phi_{2\bar2}} $$ where $\phi_{i\bar{j}}+\overline{\phi_{j\bar i}} = 0$, and let $\mathrm{U}(2)\subset \mathrm{SU}(1,2)$ be the connected subgroup on which the $\omega_i$ vanish. Then $\mathbb{CH}^2 = \mathrm{SU}(1,2)/\mathrm{U}(2)$, and the pullback of its metric to $\mathrm{SU}(1,2)$ is $\omega_1{\circ}\overline{\omega_1}+\omega_2{\circ}\overline{\omega_2}$.
Now let $\iota:G\hookrightarrow\mathrm{SU}(1,2)$ be the connected subgroup of (real) dimension $4$ such that $$ \iota^*(\gamma) = \pmatrix{ (\alpha{-}\bar{\alpha}) & \sqrt{2}(\beta{+}\alpha) & \sqrt{2}(\bar\beta{-}\bar\alpha)\\ \sqrt{2}(\bar\beta{+}\bar\alpha) & \frac12(\bar\alpha{-}\alpha)+\bar\beta-\beta & \frac12(\alpha{+}3\bar\alpha)\\ \sqrt{2}(\beta{-}\alpha) & -\frac12(\bar\alpha{+}3\alpha) & \frac12(\bar\alpha{-}\alpha)-\bar\beta+\beta} $$ where $\alpha,\bar\alpha,\beta,\bar\beta$ are linearly independent and satisfy $$ \begin{aligned} \mathrm{d}\alpha &= -\tfrac12(\alpha\wedge\beta+5\,\alpha\wedge\bar\beta-3\,\bar\alpha\wedge\beta +\bar\alpha\wedge\bar\beta)\\ \mathrm{d}\beta &= \bar\beta \wedge\beta. \end{aligned} $$ Note that $\iota^*(\omega_1{\circ}\overline{\omega_1}+\omega_2{\circ}\overline{\omega_2}) = 4(\alpha{\circ}\bar\alpha+\beta{\circ}\bar\beta)$. Meanwhile, the above structure equations imply that there exists a unique Lie group homomorphism $\rho:G\to\mathrm{SU}(1,1)$ such that $$ \rho^{-1}\mathrm{d}(\rho) = \pmatrix{\frac12(\bar\beta{-}\beta) & \bar\beta \\ \beta & \frac12(\beta{-}\bar\beta)}. $$ Let $\mathrm{U}(1)\subset\mathrm{SU}(1,1)$ be the diagonal subgroup, so that $\mathbb{CH}^1 = \mathrm{SU}(1,1)/\mathrm{U}(1)$. Then the above structure equations imply that there exists a unique smooth map $\pi:\mathbb{CH}^2\to\mathbb{CH}^1$ such that $\pi\bigl(g\mathrm{U}(2)\bigr) = \rho(g)\mathrm{U}(1)$ for all $g\in G$. Since $\rho$ pulls back the natural metric on $\mathbb{CH}^1$ to be $\beta{\circ}\bar\beta$, and since this is the natural metric on $\mathbb{H}^2$ divided by $4$ (as explained above). It follows that, $\pi$ can be viewed as a Riemannian submersion of $\mathbb{CH}^2$ onto $\mathbb{H}^2$.
That $G$, $\rho$ and $\pi$ have the properties claimed is now a matter of routine checking.