Let $\rho(A)$ denote the spectral radius of $A$. Handelman [Positive matrices and dimension groups affiliated to C∗-algebras and topological Markov chains. J. Operator Theory 6 (1981), no. 1, 55–74; MR0637001], Noutsos [On Perron-Frobenius property of matrices having some negative entries. Linear Algebra Appl. 412 (2006), no. 2-3, 132–153; MR2182957], and Johnson and Tarazaga [On matrices with Perron-Frobenius properties and some negative entries. 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 $\rho(A)$.
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]). 
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, which differs from what is established in the papers above, is not quite correct. Let $A$ be a real matrix with spectrum $\{ \lambda_1, \dots, \lambda_n \}$ and suppose that $\rho(A)$ is a positive simple eigenvalue of $A$ (for definitiness, 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 $\rho(A^q) = \lambda_1^q = \rho(A)^q$. This contradicts the Perron Frobenius theorem for positive matrices. Thus, the left and right eigenvectors corresponding to the spectral radius of an eventually positive matrix must be positive (or both negative).