Here is a general continuity type result on perturbations of the spectral radius:
Theorem. Let $A,B \in \mathbb{C}^{n \times n}$ be matrices and let $r(A)$ denote the spectral radius of $A$. For each $r > r(A)$ we set
\begin{align*}
\alpha(r) := \sup_{\lvert\lambda\rvert \ge r} \lVert (\lambda - A)^{-1} \rVert
\end{align*}
(note that $\alpha(r) < \infty$). If $\lVert B \rVert < 1/\alpha(r)$, then $r(A+B) < r$.
Proof. This is a simple consequence of the Neumann series: Let $\mu$ be a complex number of moduls $\lvert \mu \rvert \ge r$. Then $\lVert (\mu - A)^{-1} B \rVert < 1$, so the matrix
\begin{align*}
\mu - (A+B) = (\mu - A) (I - (\mu - A)^{-1} B)
\end{align*}
is invertible since $(\mu - A)$ and $I - (\mu - A)^{-1}B$ are invertible (the latter due to the Neumann series).
The following formulation of the theorem is probably a bit easier to read:
Corollary. Let $A \in \mathbb{C}^{n \times n}$. For every $r > r(A)$ there exists $\delta(r) > 0$ such that $r(A+B) < r$ for every matrix $B \in \mathbb{C}^{n \times n}$ which fulfils $\lVert B \rVert < \delta(r)$.
Proof. Take $\delta := 1/\alpha(r)$.
Remark 1. If $r(A) = 1$ and if $A$ is in addition power-bounded, i.e. $M := \sup_{n \in \mathbb{N}_0} \lVert A^n \rVert < \infty$, then we have $\alpha(r) \le M/(r - 1)$ for each $r > 1$ (this is again a simple consequence of the Neumann series representation of the resolvent). Hence, in this special case we can choose $\delta(r) = (r-1)/M$ in the corollary.
Remark 2. The above results are very rough (as indicated by their elementary proofs). There exist more precise (and quantitative) results, as for instance indicated by Federico Poloni in the comments.
