It seems that it is OK that principal $O(n)$-bundle be constructed from non-orthogonal frames. For example, $(E,p,M)$ is a vector bundle and $P$ is the frame bundle. Then smoothly at each point $x$ of $M$ choose a frame in the fiber $P_x$. Then at each point $x$ let each element of $O(n)$ act on the chosen frame $f$ at $x$ to get a set $Q_x$, glue $Q_x$ together seems to get exactly the same as the principal $O(n)$-bundle constructed from orthogonal frames, or say, they are trivially identical.
Is that right? Does the gluing process need some property special to orthogonal frame?
Here is a short answer to what I think is your main question. The reason you usually think of principal $O(n)$-bundles as having fibers consisting of the space orthonormal frames is that $O(n)$-acts simply transitively on these. Any other space on which $O(n)$-acts simply transitively has a similar interpretation as the fibers of the bundle.
As for your argument, you will not be able to "smoothly" choose a frame in the fiber $P_x$ unless your bundle was trivial to begin with. This is easy to see since such a choice would constitute a section of the (principal) frame bundle.
You can however do your construction locally in any bundle. If you take some frame $f$ and act on it with all of $O(n)$ the set you will get is just orthonormal bases with respect to some other inner product. More, precisely, one in which $f$ is orthonormal. So in some sense your idea is accomplishing is not replacing orthonormal frames with something else, it is just changing the inner product you are using to define what it means to be orthonormal.