Can a continuous map $S^2 \rightarrow S^2$ preserve orthogonality without being an isometry?

In the case that you are willing to assume that the map is once differentiable, the answer is yes.

Sketch of proof:

There is a duality between great circles on $\mathbb{S}^2$ and lines in $\mathbb{R}^3$: you can identify a great circle on $\mathbb{S}^2$ with plane in $\mathbb{R}^3$ that contains the great circle, and then identify with the line that is normal to the plane. In other words, you can write $C_v = \{ w\in \mathbb{S}^2 | w\cdot v = 0 \}$ for any $v\in\mathbb{S}^2$.

The preservation of orthogonality means, therefore, that your map $\phi$ sends great circles to great circles.

Furthermore, we observe that fixing a point $p\in \mathbb{S}^2$, we can identify its tangent directions with the collection of all great circles through it. For $\eta,\omega\in T_pM$, the angle between them can be measured by the angle between their corresponding great circles, which is the same as the angle between their corresponding dual vectors.

So: if $\phi$ is $C^1$, the differential $d\phi$ defines a linear map between the tangent spaces. That $\phi$ preserves orthogonal directions now implies that $d\phi$ preserves orthogonal directions. Hence $d\phi$ must be conformal! (Since it is linear and preserves orthogonality.) So $\phi$ is a conformal map of the sphere. But recall back that $\phi$ preserves all great circles--any conformal automorphism of the sphere that preserves all great circles must be an isometry.