The title might be slightly misleading (in that I don't need the flow to contract $M$ in finite time). What I'm asking is this:
If $M$ is a contractible smooth $n$-manifold and $p \in M$, is there a smooth vector field $X$ on $M$ s.t. its flow $\Phi^X: [0, \infty) \times M \to M$ is defined for all positive times and for all neighborhoods $U$ of $p$ there is a $T$ s.t. $$ \Phi^X_t(M) \subset U \quad\text{for}\quad t > T \tag{1} $$
This should be equivalent to $\Phi^X$ extending to a continuous map $[0, \infty] \times M \to M$ by $\Phi^X_\infty(x) = p$.
This is obviously true for $\mathbb{R}^n$, where we can take $X(x) = -x$.
I don't really understand the Whitehead manifold (a contractible 3-fold that is not homoemorphic to $\mathbb{R}^3$), so I haven't checked that.
Edit: I just noticed that picking $U$ to be a coordinate ball (1) shows that $M$ must embed into $\mathbb{R}^n$ (via $\Phi^X_T$). Is it possible that this gives a contradiction when $M$ is not $\mathbb{R}^n$?
Background: It seems to me that this is (implicitly) used in Voisin's Hodge Theory and Complex Algebraic Geometry when proving Ehresmann's theorem (Thm 9.3). The last paragraph of the proof states:
Here $U$ is a small neighborhood of an arbitrary point on the manifold $B$.