Let $M$ be an $n$-dimensional topological closed manifold. Suppose $K$ is a compact subset of $M$ which is contractible in the sense that there exists a continuous map $F:K \times [0,1] \to M$ with $F(\cdot,0)=id$ and $F(\cdot, 1)=q \in M$.
Can we find an open neighborhood $U$ of $K$ such that $U$ is homeomorphic to $\mathbb R^n$?
This is not true. Let K be one component (it doesn't matter which one) of the Whitehead link, which has two components. Then K is contractible in the complement M of the other component. But K is not contained in a 3-ball in M. This can be seen in many ways; for instance if K were contained in a 3-ball, then each of its lifts to the universal cover of M would have trivial linking numbers. But you can readily draw the picture of the cover and see that some of those linking numbers are $\pm 1$.
No. Let $M$ be $S^1\times S^2$ and let $K$ be homeomorphic to $S^1$ and chosen such that in the universal cover $\tilde M=\mathbb R\times S^2$ there are two liftings $K_1$ and $K_2$ of $K$ which are linked together. If $U$ existed then these would be contained in two disjoint liftings $U_1$ and $U_2$ of $U$, therefore unlinked.
We know that the answer to the original posted question is negative. But I think it might be worth giving an affirmative answer to the case when $n=2$ so one can see the obstructions to generalization of higher dimensional cases, for instance, some obstructions exploited by Siebenmann as Ryan Budney suggested. The proof is based on my previous comments. I believe Tom Goodwillie has a proof as well.
Lemma. Let $K$ denote a compact subset of an ANR $X$. Then $K$ is contractible in $X$ iff some nbhd $U$ of $K$ is contractible in $X$.
Proof. Let $F_t$ denote a contraction of $K$ in $X$. Define a map $f:A\to X$ on $A=(K\times[0,1])∪(X×\{0,1\})$ of $X\times[0,1]$ as $f(<x,0>)=x$ and $f(<x,1>)=F_1(K)=point$ for $x\in X$ and as $f(<k,t>)=F_t(k)$ for $k\in K$. Since $X$ is an ANR, $f$ extends to a map $F′:V\to X$ defined on a nbhd $V$ of $A$ in $X×[0,1]$. Then $K$ has a nbhd $U$ with $U×[0,1]$ in $V$, and the restriction on $U×[0,1]$ is the desired contraction. The other implication is obvious.
Claim. Let $K$ be a compact subset of a 2-manifold $M$. If $K$ is contractible in $M$, then there exists a nbhd $U$ of $K$ which is homeomorphic to $\mathbb{R}^2$.
Sketch of the proof. The argument is mainly focused on the case $M = \mathbb{R}^2$. The general case will follow by lifting to the universal cover of $M$, which is topologically $\mathbb{R}^2$ or $S^2$. By Lemma above, there exists a nbhd $U$ which contracts in $\mathbb{R}^2$. Build a compact 2-manifold with boundary $H$ in $\mathbb{R}^2$ such that $K \subset Int H \subset H \subset U$, $Int H$ is connected and $\partial H$ is polygonal. If $\partial H$ is connected, Jordan-Schonflies theorem implies that $H$ is a 2-cell. Otherwise, one may cut away at $H$ until it is connected. This is not too hard to do since $H - K$ is connected. This follows from Hurewicz-Wallman[P.100] saying a compact subset of $\mathbb{R}^n$ separates $\mathbb{R}^n$ iff it admits a map to $S^{n-1}$ that is not nullhomotopic. Therefore, we can run a polygonal arc through $H - K$ from one boundary component of $\partial H$ to another. Thicken the arc to a disk $D$ so that $H'=Cl(H -D)$ is a compact 2-manifold with boundary. Keep cutting away like this until a 2-cell appears.
