This is true. I needed to use some higher tech than I would have expected to prove it; I'd be curious to hear a simpler solution.
Lemma: Let $M$ be a smooth manifold, $X$ a smooth vector field on $M$, $\theta$ the flow along $X$ and $x$ a point of $M$. Then there is a positive number $\epsilon$, and open sets $U \supset V \ni x$, such that, for $t \in (0, \epsilon)$, the flow $\theta_t$ takes $V$ into $U$, and the only fixed points of $\theta_t$ are the zeroes of $X$.
Proof: Since the statement is local, we may immediately assume that $X = \mathbb{R}^n$. We will fix a Euclidean norm on $\mathbb{R}^n$, so we may talk about arc-lengths, surface areas and so forth.
Take $U$ to be an open ball around $x$ (of finite radius). Choose $\epsilon'$ small enough that flowing from $x$ for time $\epsilon'$ stays within $U$. Choose $V$ a small enough ball around $U$ that flowing by time $\epsilon'$ keeps $V$ within $U$.
Let $K$ be a bound for $|\nabla \times X|$. (If $n=2$ or $3$, you presumably know what this means. In general, I mean to use the inner product on $\mathbb{R}^n$ to turn $X$ into a $1$-form, take $d$ of that $1$-form and then use the induced norm on $\bigwedge^2 \mathbb{R}^n$. I can't tell from your writing whether you are happy with this sort of manipulation -- if not, just think about the curl you are used to.) Our $\epsilon$ will be $\min(\epsilon', 4 \pi K^{-1})$.
Consider a nontrivial closed flow line, $\gamma$, in $U$. Let $T$ be the time taken to transverse $\gamma$; we will show $T>\epsilon$. Let $\sigma$ be the disc of minimal area with boundary $\gamma$. Let $L$ be the length of $\gamma$ and let $A$ be the area of $\sigma$.
By the Cauchy-Schwarz inequality,
$$\int_{\gamma} |X|^{-1} ds \cdot \int_{\gamma} |X| ds \geq \left( \int_{\gamma} ds \right)^2 = L^2.$$
Here the integrals are with respect to arc-length.
Now, $\int_{\gamma} |X|^{-1} ds = T$. (The time to travel a path is the integral, over the length of the path, of the inverse speed. If you run one $7$-minute mile, and one $9$-minute mile, it's going to take you $16$ minutes to run two miles.)
Since $\gamma$ is a flow line for $X$, we see $\int_{\gamma} |X| ds = \int X \cdot ds$, the line integral of $X$ along $\gamma$. By Green's theorem, this is the same as $\int_{\sigma} \nabla \times X$, which is $\leq K A$.
Putting it all together,
$$T \cdot (KA) \geq L^2.$$
Now, $\sigma$ is the minimal surface with boundary $\gamma$. By a result of Carleman, the isoperimetric inequality $A \leq L^2/(4 \pi)$ holds for $\sigma$ and $\gamma$. The best online refence I could find for the result of Carleman is this paper of Choe; Carleman's result is discussed in the third paragraph.
So
$$T \cdot K \cdot L^2/(4 \pi) \geq L^2$$
and we deduce that
$$T \geq 4 \pi/K \geq \epsilon$$
as desired. This proves the lemma.
Now that we have the lemma, we can find such a $(V,U, \epsilon)$ for every $x \in M$. We can find finitely many $V$'s with cover $X$ and, taking the minimum of the finitely many $\epsilon$'s, we find that $X$ has no nontrivial cycles with length $< \epsilon$. QED