This is my comment above slightly expanded. Let's focus on $2\times 2$ matrices for convenience and let $A(z)$ be entire with $\det A=1$ (divide through by a holomorphic square root of $\det A$ if this doesn't hold initially, and recall that $\sqrt{\det A(z)}\not= 0$ does have an entire scalar logarithm).
If $A(z)=e^{B(z)}$, then $e^{\textrm{tr}\: B}=1$, so $\textrm{tr}\: B(z)\equiv 2\pi in$, and by replacing $B$ with $B-i\pi n$ (and perhaps $A$ with $-A$), we can then also assume that $\textrm{tr}\: B=0$.
The eigenvalues of $A$ are $\lambda(z)=T(z)/2 \pm (1/2)\sqrt{T^2(z)-4}$, $T(z)=\textrm{tr}\: A(z)$. These are typically not entire, but only have Puiseux series expansions about the points $z_0$ with $T(z_0)=\pm 2$. As a consequence, $$ \det B(z)=\log\lambda\cdot\log 1/\lambda =-\log^2\lambda(z) $$ will be prevented from being holomorphic near such points $z_0$, and hence there is no entire matrix logarithm.
Concrete examples where this happens are now easy to find. For example $$ A(z) = \begin{pmatrix} 1 & z\\ 1 & 1+z\end{pmatrix} $$ works (and the problematic point is $z_0=0$), or of course the example given by Robert, which in normalized form is $$ A(z) = \begin{pmatrix} e^{z/2} & 0\\ ze^{-z/2} & e^{-z/2} \end{pmatrix} , $$ and here $\textrm{tr}\: A=\pm 2$ when $e^{z/2}=\pm 1$, consistent with Robert's calculation.