This answer gives some insight on *eventually positive* matrices, which differs from the original question regarding *power nonnegative* matrices. 

Let $\rho(A)$ denote the spectral radius of $A$. Handelman [J. Operator Theory 6 (1981), no. 1, 55–74; MR0637001], Noutsos [Linear Algebra Appl. 412 (2006), no. 2-3, 132–153; MR2182957], and Johnson and Tarazaga [Positivity 8 (2004), no. 4, 327–338; MR2117663] showed that a real matrix $A$ is *eventually positive* if and only if $\rho(A)$ is a positive simple eigenvalue satisfying 
$$
|\lambda| < \rho(A)
$$
for every $\lambda \in \sigma(A)$, and there are positive left and right eigenvectors $u$ and $v$ corresponding to $\rho(A)$ (this is known as the *strong Perron Frobenius property*).

It is also known that *power index of $A$*, which is the smallest positive integer $q$ such that $A^k$ is positive for all $k \ge q$ can be arbitrarily large. This is because, under very mild assumptions, arbitrarily large roots of eventually positive matrices remain eventually positive (see McDonald et al. [Matrix roots of eventually positive matrices. Linear Algebra Appl. 456 (2014), 122–137; MR3223894]). [![thm3.5][1]][1]

Another important work on eventually nonnegative matrices is by McDonald and Zaslavsky [A characterization of Jordan canonical forms which are similar to eventually nonnegative matrices with the properties of nonnegative matrices. Linear Algebra Appl. 372 (2003), 253–285; MR1999150]. 

It is, however, known that if $A$ is a *primitive matrix*, then $n^2 - 2n+2$ is a sharp upper bound on the *index of primitivity* (see Chapter 8 of *Matrix Analysis* by Horn & Johnson). 

EDIT: The answer above by R. Israel is not quite correct.    

Let $A$ be a $n$-by-$n$ matrix with spectrum $\{ \lambda_1, \dots, \lambda_n \}$ and suppose that $\rho(A)$ is a simple eigenvalue of $A$ (for definiteness, assume $\lambda_1 = \rho(A)$). For contradiction, assume that $v$ is a totally nonzero eigenvector containing some negative entries. If $A^q > 0$, then $A^q v = \lambda_1^q v$. However, since 
$\lambda_i < \lambda_1$ for every $i$, and $\{ \lambda_1^q,\dots,\lambda_n^q \}$ is the spectrum of $A$, it follows that $0 < \rho(A^q) = \lambda_1^q$. This contradicts the Perron Frobenius theorem for positive matrices. Thus, the left and right eigenvectors corresponding to the spectral radius of a power positive matrix must be positive (or both negative).  

  

  [1]: https://i.sstatic.net/DwiO4.png