This is not possible; sorry for just posting a sketch but I am not on MO :-(

For instance a generic geodesic triangle in the hyperbolic plane does not embed in a finite product of trees.

Suppose you have a finite product $X$ of uniquely geodesic metric spaces $X_n$ with the "pythagorean" distance on the product. Then it is uniquely geodesic and any geodesic segment $\sigma\colon[a,b]\to X$ is of the form $\sigma(t)=(\sigma_n(a_n t))_n$ where $a_n\geq 0$ is a sequence with $\sum a_n^2 = 1$ and each $\sigma_n$ is a geodesic in $X_n$. This characterisation is not entirely trivial, but very well-known (see e.g. Bridson--Haefliger).

Now in the case at hand you would deduce that the distance between two geodesics issuing from a common point would be a piecewise affine function, which it definitely isn't for hyperbolic geometry.

The fact about geodesics extends to infinite sums, and even to integrals (see e.g. Proposition 44 in http://dx.doi.org/10.1090/S0894-0347-06-00525-X )

The latter part of the argument above (piecewise affine) does not extend as is to this "integral" setting though.

This is not possible; sorry for just posting a sketch but I am not on MO :-(

For instance a generic geodesic triangle in the hyperbolic plane does not embed in a finite product of trees.

Suppose you have a finite product $X$ of uniquely geodesic metric spaces $X_n$ with the "pythagorean" distance on the product. Then it is uniquely geodesic and any geodesic segment $\sigma\colon[a,b]\to X$ is of the form $\sigma(t)=(\sigma_n(a_n t))_n$ where $a_n\geq 0$ is a sequence with $\sum a_n^2 = 1$ and each $\sigma_n$ is a geodesic in $X_n$. This characterisation is not entirely trivial, but very well-known (see e.g. Bridson--Haefliger).

Now in the case at hand you would deduce that the distance between two geodesics issuing from a common point would be a piecewise affine function, which it definitely isn't for hyperbolic geometry.

The fact about geodesics extends to infinite sums, and even to integrals (see e.g. Proposition 44 in https://dx.doi.org/10.1090/S0894-0347-06-00525-X )

The latter part of the argument above (piecewise affine) does not extend as is to this "integral" setting though.