Continuity of eigenvectors

Yes. Let the size of your matrix be $n$. Your condition implies that there is an $n-1times n-1$ submatrix whose determinant is not identically equal to $0$. Assume without loss of generality that this is the submatrix formed by the first $n-1$ rows and columns. Then we can set $u_n=1$ and find a vector $u(z)$ such that $M(z)u(z)=0$ by solving the system of $n-1$ linear equations in $n-1$ variables. This requires only arithmetic operations on the entries of $M$, so $u(z)$ will be meromorphic on $mathbb{C}$. Let $D$ be the divisor of poles of $u$. According to a theorem of Weierstrass there is an entire function $f$ having zeros at $D$. Then $v(z)=f(z)u(z)$ is a solution to your problem, which is not only continuous but holomorphic.