There is a one-dimensional counterexample: Consider the analytic semigroup $z \mapsto e^{iz}$. This semigroup is bounded on the non-negative real line, but it is not bounded on any sector $\Delta_\delta$.
EDIT in response to the comments: One can "modify" each analytic semigroup to obtain the following boundedness property:
Let $0 < \delta' < \delta \le \frac{\pi}{2}$ and let $(T(z))_{z \in \Delta_\delta}$ be an analytic semigroup. Then there exists a real number $\omega \ge 0$ such that the rescaled semigroup $(e^{-\omega z} T(z))_{z \in \Delta_\delta}$ is bounded on the sector $\Delta_{\delta'}$.
This can be seen as follows: Let $\lambda$ be the complex number with modulus $1$ and argument $\delta'$. Then $(T(t\lambda))_{t \in [0,\infty)}$ and $(T(t\overline{\lambda}))_{t \in [0,\infty)}$ are $C_0$-semigroups, so there exists a number $\eta \ge 0$ and a number $M \ge 1$ such that $\|T(t\lambda)\| \le Me^{\eta t}$ and $T(t\overline{\lambda}) \le M e^{t\eta}$ for all $t \in [0,\infty)$.
Choose $\omega = \frac{\eta}{\operatorname{Re} \lambda}$. Let $z$ be a number on the boundary of the sector $\Delta_{\delta'}$. Then $z$ is of the form $z = t \lambda$ or $z = t\overline{\lambda}$ for a real number $t \ge 0$. Hence, $\|e^{-\omega z} T(z)\| \le e^{-\omega t \operatorname{Re} \lambda} M e^{\eta t} = M$. Since the sector $\Delta_{\delta'}$ is contained in the convex hull its boundary, we conclude that $\|e^{-\omega z} T(z)\| \le M^2$ for all $z \in \Delta_{\delta'}$. Hence, the semigroup $(e^{-\omega z}T(z))_{z \in \Delta_\delta}$ is bounded on the sector $\Delta_{\delta'}$.
Note however that $M$ and $\omega$ might both depend on $\delta'$. The one-dimensional counterexample above shows that we cannot choose $\omega$ independently of $\delta'$ in general.