Here is an outline (I do not need the geodesic property, only properness of your metric space). If you are interested in metric geometry, you should be able to fill in the details.

Definition. A **separated net** in a metric space $(X,d)$ is a subset $N\subset X$ such that for some $r>0, \epsilon>0$ the following hold:

(a) $\bigcup_{x\in N} B(x,r)=X$ (i.e. $N$ is an net).

(b) $B(x,\epsilon)\cap B(y,\epsilon)=\emptyset$ for all distinct points $x, y\in N$ (i.e. $N$ is separated).

I leave it to you as an exercise to prove that:

(1) A separated net exists in every metric space.

(2) If $(X,d)$ is proper, then every ball $B(x,R)\subset X$ contains only finitely many elements of $N$.

(3) The inclusion map $\iota: (N,d|_N)\to (X,d)$ is a quasi-isometry.

The key is to prove that for every separated net $N\subset X$ in a proper metric space, there exists a Borel map $q: X\to N$ which is a quasi-inverse to the quasi-isometry $\iota$. Once you have this, then for every quasi-isometric embedding
$$
f: (X,d)\to (X',d')
$$
you take $g:= f\circ q$, a new quasi-isometric embedding which is Borel and is at a finite distance from $f$.

Now, I will define $q$. Given a separated net $N\subset X$, consider the corresponding **Voronoi tiling** of $X$ corresponding to $N$. The (closed) tile corresponding to a point $x\in N$ will be denoted $V_x$:
$$
V_x=\{y\in X: d(x,y)\le d(z,y) \quad \forall z\in N\}.
$$
In view of properness of $(X,d)$, each point in $X$ belongs only to finitely many tiles. Moreover, the diameter of each tile is $\le 2r$.

Each tile is a Borel subset and so are the intersections of the tiles
$$
V_{x_1,...,x_k}=V_{x_1}\cap .... \cap V_{x_k},
$$
(the "closed strata" of the tiling) as well as the following subsets, "open strata" of the tiling:
$$
V^{'}_{x_1,...,x_k}:= V_{x_1,...,x_k}- \bigcup_{x\in N - \{x_1,...,x_k\}} V_x.
$$
For instance, for $x_1\in N$,
$$
V^{'}_{x_1}= \{y\in X: d(x_1,y)< d(z,y) \quad \forall z\in N-\{x_1\}\}.
$$

Then $X$ is a disjoint union of the open strata as above. I will order arbitrarily all the finite subsets $\{x_1,...,x_k\}$ of $N$ for which
$V^{'}_{x_1,...,x_k}$ is nonempty and let $x_1$ denote the smallest element of this finite subset.

Define the map
$$
q: X\to N, q(x)=x_1,
$$
where $x\in V^{'}_{x_1,...,x_k}$.

(4) I leave it to you to check that $q$ is a Borel quasi-isometry as required.

## Best Answer

For a counterexample, take $(S,d_1)$ to be the Euclidean plane and $(S',d_2)$ to be the union of the integer coordinate lines in the Euclidean plane: $$S' = (\mathbb R \times \mathbb Z) \cup (\mathbb Z \times \mathbb R) $$

For a counterexample with complete Riemannian surfaces, start with the same $(S,d_1)$. For $(S',d_2)$, start with the Euclidean plane $S$, consider each integer coordinate square $[m-1,m] \times [n-1,n]$, and then for each $m,n$ replace that square by its connected sum with a torus.