Lower bound for the speed convergence of the spectrum

Question

Let $A$ be a (complex unital) Banach algebra and $a \in A$ with spectrum $\sigma(a) = \{ 0 \}$ ($a$ is quasinilpotent). For $b \in A$ and $\varepsilon > 0$ consider the linear perturbation $a + \varepsilon b$. Then in $\sigma(a + \varepsilon b)$ the spectral point $0$ of $a$ is split into spectral sets, so that in particular $\sigma(a + \varepsilon b) \setminus \{ 0 \}$ is a spectral set (and if $0 \in \sigma(a + \varepsilon b)$ then $\{ 0 \}$ is also a spectral set of $a + \varepsilon b$). Since $\sigma(a) = \{ 0 \}$ is finite one can show that $\sigma(a + \varepsilon b)$ converges to $\sigma(a)$. I am interested in a lower bound for the speed of this convergence in the following sense: is it true that there exists $K > 0$ such that for each $\varepsilon$ small enough the open ball $B_{K\varepsilon}(0) \subseteq \mathbb{C}$ around $0$ with radius $K\varepsilon$ does not contain any spectral point of $\sigma(a + \varepsilon b)$ except possibly $0$, i.e. $B_{K\varepsilon} \cap (\sigma(a + \varepsilon b) \setminus \{ 0 \}) = \emptyset$? Equivalently, does it hold for the distance that $d(\sigma(a + \varepsilon b) \setminus \{ 0 \}, \{ 0 \}) \geq K \varepsilon$ for each $\varepsilon$ small enough?