In addition to DBM's (totally correct) answer above, I realized that there's probably a much simpler answer. If I'm wrong, hopefully someone will set me right.

Let $\mathcal O$ be an open neighborhood in $\mathbb R^n$ with compact closure. Consider the family of differential equations: $$ 0 = \ddot\gamma + \epsilon \\, db \cdot \dot \gamma + \epsilon^2 \\,dc \quad\quad ({\rm EOM}\_\epsilon)$$ The solutions to $\rm (EOM\_0)$ are just straight lines. For each $\epsilon$, consider the flow $\phi\_\epsilon: {\rm T}\mathcal O \to \mathbb R^{2n}$, which sends $(v,q)$ to $\bigl(\varphi\_\epsilon(v,q),q\bigr)$, where $\varphi_\epsilon(v,q) \in \mathbb R^n = \gamma\_\epsilon(1)$, where $\gamma\_\epsilon$ solves $\rm (EOM\_\epsilon)$ with initial conditions $\bigl(\dot\gamma(0),\gamma(0)\bigr) = (v,q)$. By the standard results from ODEs, $\phi_\epsilon$ is smooth when it's defined, and depends smoothly on $\epsilon$.

But the closure of $\mathcal O$ is compact, so $\phi_\epsilon$ is defined for sufficiently small $\epsilon$ depending only on $\mathcal O$. Moreover, $\phi\_0(v,q) = (q+v,q)$ is one-to-one, and $\phi\_\epsilon$ is too for $\epsilon$ sufficiently small depending on $\mathcal O$. But the flow by time $1$ for $\rm (EOM\_\epsilon)$ is, up to a linear reparameterization, equivalent to the flow by time $\epsilon$ for $\rm (EOM\_1) = (EOM)$.

Thus we have a solution to part 1. of the conjecture/question. And part 2. is essentially obvious, because this family contains all "low energy" paths that have both endpoints in $\mathcal O$. So this doesn't quite do 2. as stated, but replacing $\mathcal O$ by $\mathcal O_0 \subseteq \overline{\mathcal O_0} \subseteq \mathcal O_1 \subseteq \overline{\mathcal O_1}$ compact does the trick.

In addition to DBM's (totally correct) answer above, I realized that there's probably a much simpler answer. If I'm wrong, hopefully someone will set me right.

Let $\mathcal O$ be an open neighborhood in $\mathbb R^n$ with compact closure. Consider the family of differential equations: $$ 0 = \ddot\gamma + \epsilon \, db \cdot \dot \gamma + \epsilon^2 \,dc \quad\quad ({\rm EOM}_\epsilon)$$ The solutions to $\rm (EOM_0)$ are just straight lines. For each $\epsilon$, consider the flow $\phi_\epsilon: {\rm T}\mathcal O \to \mathbb R^{2n}$, which sends $(v,q)$ to $\bigl(\varphi_\epsilon(v,q),q\bigr)$, where $\varphi_\epsilon(v,q) \in \mathbb R^n = \gamma_\epsilon(1)$, where $\gamma_\epsilon$ solves $\rm (EOM_\epsilon)$ with initial conditions $\bigl(\dot\gamma(0),\gamma(0)\bigr) = (v,q)$. By the standard results from ODEs, $\phi_\epsilon$ is smooth when it's defined, and depends smoothly on $\epsilon$.

But the closure of $\mathcal O$ is compact, so $\phi_\epsilon$ is defined for sufficiently small $\epsilon$ depending only on $\mathcal O$. Moreover, $\phi_0(v,q) = (q+v,q)$ is one-to-one, and $\phi_\epsilon$ is too for $\epsilon$ sufficiently small depending on $\mathcal O$. But the flow by time $1$ for $\rm (EOM_\epsilon)$ is, up to a linear reparameterization, equivalent to the flow by time $\epsilon$ for $\rm (EOM_1) = (EOM)$.

Thus we have a solution to part 1. of the conjecture/question. And part 2. is essentially obvious, because this family contains all "low energy" paths that have both endpoints in $\mathcal O$. So this doesn't quite do 2. as stated, but replacing $\mathcal O$ by $\mathcal O_0 \subseteq \overline{\mathcal O_0} \subseteq \mathcal O_1 \subseteq \overline{\mathcal O_1}$ compact does the trick.