Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature
I'm not sure what you mean by "determining". One natural notion of equivalence is for two surfaces to be related by an ambient isometry (a euclidean motion).
A basic result is that two surfaces in $\mathbb{R}^3$ are related by an isometry of $\mathbb{R}^3$ if and only if their first and second fundamental forms agree.
A weaker condition is that of isometry. Two surfaces are said to be isometric if their first fundamental forms agree. Gauss's Theorema Egregium says that isometric surfaces have the same Gaussian curvature, but the converse is not true: there are examples of surfaces with the same Gaussian curvature, but which are not isometric.
In dimension $\geq 4$ Kulkarni in his paper Curvature and Metric showed that a diffeomorphism which preserves the sectional curvature is an isometry, except possibly in the case of constant sectional curvature. In dimension $\leq 3$ there are counterexamples which are mentioned in his paper.
The associated family of a minimal surface gives a tangible counterexample. The Weierstrass representation lets you cook up a conformally parameterized minimal surface from a meromorphic pair $f \sqrt{dz}$, $g \sqrt{dz}$.
The parameterization is then given by $$F(x,y) = Re\int_0^{x+iy} (f^2 - g^2, i(f^2 + g^2), 2 fg) ~dz$$
The normal map of $F$ can be obtained by thinking of $g/f$ as a map to the Riemann sphere, and the metric induced by $F$ is just $4(|f|^2 + |g|^2)^2 |dz|^2$. From this data, you can cook up the Gauss and mean curvatures, and it happens to be true that if $f \sqrt{dz}$ and $g \sqrt{dz}$ are meromorphic, you get a minimal surface.
But then consider what happens if you multiply both $f$ and $g$ by $e^{i \theta}$ --- the normal map and the metric are both unchanged, and $e^{i\theta} f \sqrt{dz}, e^{i\theta} g \sqrt{dz}$ are still quite meromorphic, so you get a new minimal surface which is isometric to your old one. This means you have made a new surface whose principal curvatures agree with your old one!
I think the moral here is that even knowing the metric and the complete set of principal curvatures isn't enough to reconstruct a surface --- the curvature directions are also vital data.
To see all this in action, here is a video with strange music showing the helicoid transforming into the catenoid, which starts with the Weierstrass data for the catenoid and then multiplies by $e^{i \theta}$, with $\theta$ increasing as the movie progresses. Every one of the surfaces is isometric to the catenoid! But they do have different second fundamental forms.
Insted of Q2, I will answer the following question:
Are there higher-dimensional generalizations, determining a submanifold in $\mathbb R^q$.
Yes, there are some analogs, but I am sure you do not need them.
They work well for $n$-dimensional submanifolds in $\mathbb R^{n{\cdot}(n+3)/2}$. (curves in $\mathbb R^2$, surfaces in $\mathbb R^5$ and so on). Instead of natural parametrization you remember metric tensor $g$; which is a degree 2 homogeneous polynomial on the tangent space. Instead of curvature you remember the following degree 4 homogeneous polynomial $h(X)=|s(X,X)|^2$, where $s\colon T\times T\to N$ is the second fundamental form (for two tangent vectors $X$ and $Y$ the value $s(X,Y)$ is a normal vector).
The proofs are the same as Frenet–Serret formulas. You can find it in Spivak's book.