{\bf Counterexample.} (But look at my comment above please.) Take your $B$ dead zero: no magnetic field, or friction (however you are thinking of it). Your force field is now pure potential. Your inital velocities $\nu$ are all zero. Take $C = (1/2)|q|^2$ -- a harmonic oscillator --so your "EOM" is $\ddot q = - q$. Take your "base" $q \ne 0$ from your question/conjecture. Conservation of energy asserts that any $q_1 = \gamma(t)$ connected to such a $q$ by the classical path $\gamma(t)$ having initial condition $(q, \nu) = (q,0)$ satisfies $|q_1|^2 \le |q|^2$. Indeed $|\gamma(t)|^2 \le |q|^2$ with equality if and only if $t$ is an integral multiple of $\pi$. In particular for all sufficiently small time we have $|\gamma(t)| < |q|$. (You have to wait a time $2 \pi$ to get back to $q$. ) Thus if you really want your time intervals small of type $(0, \epsilon)$ with $\epsilon$ small you are screwed! You cannot have both $\mathcal O_0$ of $\mathcal O_1$ containing $q$, since $\mathcal O_0 \times \mathcal O_1$ cannot contain any point along the diagonal.
I may have gotten $q$ and $q_1$ reversed relative to your labellings of your question/conjecture, but the same trick still works. The guts of the matter is that a ball of initial conditions of the form $(q, \nu) = (q,0)$ shrinks in $q$-space under the oscillator flow: the force is attractive, after all!
The same trick is bound to work for non-zero $B$.
You might be able to `save' your conjecture by rephrasing, eg. not insisting that $\mathcal O_0 \times \mathcal O_1$ intersect the diagonal, but keep the oscillator in mind, and perhaps say more clearly where you are really headed in posting this question/conjecture.
