The answer to your question is positive for $n>1$, and this follow just from the fact that a holomorphic function defined on the complement to a pseudoconvex domain can be always extended to the domain for $n>1$. For $n=1$ the statement not true (as Pietro Majer says correctly says it). There is a reference now given in a comment by Margaret Friedland that justifies this answer.
In the case that you consider there is a holomorphic map from
$M\setminus K$ to $\mathbb C^n\setminus \bar B_1$, i.e. you have $n$ holomorphic functions
on $M\setminus K$. Each of these functions can be extended to $M$ provided $n>1$ since $K$ is pseudoconvex in $M$. So you get a proper holomorphic map from $M$ to $\mathbb C^n$. Moreover this map is birational (or of degree one in other words), so $M$ is a contractible topological space only if the map is an isomorphism (otherwise there will be some exceptional divisors on $M$ that will be contracted to points by the map and so the topology of $M$ will be non-trivial).
Here is the mathscient citation for the reference given by Margaret Friedland
"The authors prove the following: If $M$ is a ﬁnite complex manifold with
connected boundary $bM$ such that the Levi form has one positive eigenvalue everywhere on $bM$, then every function on $bM$
which satisﬁes the “tangential Cauchy-Riemann equations” on $bM$ has a holomorphic extension to the whole of $M$."
Note that the boundary of a ball in $\mathbb C^n$ has positive Levi form (for $n>1$) and the "tangential Cauchy-Riemann" equation is automatically satisfied provided the function is defined and holomorphic in a neighbourhood of $bM$. Clearly we can assume the later in our case.