As already noted by several users in this comments, something seems to be a bit odd with the question due to the behaviour of the resolvent close to the spectrum. Here are a few details about what is true and what is not:

1) If $(z_n)$ is a sequence in $\mathbb{C} \setminus \sigma(T)$ such that $\|R(z_n,T)\| \to 0$, then we necessarily have $\operatorname{dist}(z_n, \sigma(C)) \to \infty$; this follows from the inequality \begin{align*} \|R(z,T)\| \ge \frac{1}{\operatorname{dist}(z, T)} \qquad (*) \end{align*} which is true for all $z\in \mathbb{C} \setminus \sigma(T)$. [To also cover the case $\sigma(T) = \emptyset$ in this formula, we define the distance of any number from the empty set to be $\infty$.]

Thus, it does not suffice to consider only numbers $z$ which are bounded away from the spectrum and tend to infinity; one has to assume that the distance of $z$ to the spectrum also tends to infinity in order to have any chance for the resolvent to tend to $0$.

2) If, on the other hand, the question is whether there exists a Banach space $X$ and a closed non-bounded operator $T$ such that $\|R(z_n,T)\| \to 0$ for any sequence $(z_n) \subseteq \mathbb{C} \setminus \sigma(T)$ which satisfies $|z_n| \to \infty$ and $\operatorname{dist}(z_n,\sigma(T)) \to \infty$, then the answer is yes:

Just choose $X$ as an infinite dimensional Hilbert space and let $H$ be your favourite non-bounded self-adjoint operator on $X$. (It follows from the spectral theorem that we have equality in $(*)$ for self-adjoint operators).

3) The question probably becomes a bit more interesting if we consider the following variant: If $T$ is a closed operator on a Banach space $X$ such that $\sigma(T)$ is bounded and such that $\|R(z,T)\| \to 0$ as $|z| \to \infty$, does it follow that $T$ is bounded?

The answer is yes (here, by ``bounded'' I mean not only that $T$ is continuous, but also that $D(T) = X$).

Proof. Since $\sigma(T)$ is bounded, we can compute the spectral projection corresponding to the entire spectrum by integrating that resolvent over a path that encloses $\sigma(T)$. This yields a decomposition of $X$ into two closed subspaces, $X = V \oplus W$, such that $T$ splits along this decomposition into a bounded operator $T_V$ on $V$ and a closed operator $T_W$, with empty spectrum, on $W$.

The resolvent of $T_W$ is an analytic function defined everywhere on $\mathbb{C}$ and by assumption the resolvent of $T_W$ decays to $0$ as $|z| \to \infty$. Hence, the resolvent of $T_V$ is constant by Liouville's theorem and thus, it is constantly $0$. But each resolvent operator has to be injective, so $W = 0$. Thus, $X = V$ and $T$ is equal to the bounded operator $T_V$.

As already noted by several users in this comments, something seems to be a bit odd with the question due to the behaviour of the resolvent close to the spectrum. Here are a few details about what is true and what is not:

1) If $(z_n)$ is a sequence in $\mathbb{C} \setminus \sigma(T)$ such that $\|R(z_n,T)\| \to 0$, then we necessarily have $\operatorname{dist}(z_n, \sigma(C)) \to \infty$; this follows from the inequality \begin{align*} \|R(z,T)\| \ge \frac{1}{\operatorname{dist}(z, T)} \qquad (*) \end{align*} which is true for all $z\in \mathbb{C} \setminus \sigma(T)$. [To also cover the case $\sigma(T) = \emptyset$ in this formula, we define the distance of any number from the empty set to be $\infty$.]

Thus, it does not suffice to consider only numbers $z$ which are bounded away from the spectrum and tend to infinity; one has to assume that the distance of $z$ to the spectrum also tends to infinity in order to have any chance for the resolvent to tend to $0$.

2) If, on the other hand, the question is whether there exists a Banach space $X$ and a closed non-bounded operator $T$ such that $\|R(z_n,T)\| \to 0$ for any sequence $(z_n) \subseteq \mathbb{C} \setminus \sigma(T)$ which satisfies $|z_n| \to \infty$ and $\operatorname{dist}(z_n,\sigma(T)) \to \infty$, then the answer is yes:

Just choose $X$ as an infinite dimensional Hilbert space and let $H$ be your favourite non-bounded self-adjoint operator on $X$. (It follows from the spectral theorem that we have equality in $(*)$ for self-adjoint operators).

3) The question probably becomes a bit more interesting if we consider the following variant: If $T$ is a closed operator on a Banach space $X$ such that $\sigma(T)$ is bounded and such that $\|R(z,T)\| \to 0$ as $|z| \to \infty$, does it follow that $T$ is bounded?

The answer is yes (here, by ``bounded'' I mean not only that $T$ is continuous, but also that $D(T) = X$).

Proof. Since $\sigma(T)$ is bounded, we can compute the spectral projection corresponding to the entire spectrum by integrating that resolvent over a path that encloses $\sigma(T)$. This yields a decomposition of $X$ into two closed subspaces, $X = V \oplus W$, such that $T$ splits along this decomposition into a bounded operator $T_V$ on $V$ and a closed operator $T_W$, with empty spectrum, on $W$.

The resolvent of $T_W$ is an analytic function defined everywhere on $\mathbb{C}$ and by assumption the resolvent of $T_W$ decays to $0$ as $|z| \to \infty$. Hence, the resolvent of $T_V$ is constant by Liouville's theorem and thus, it is constantly $0$. But each resolvent operator has to be injective, so $W = 0$. Thus, $X = V$ and $T$ is equal to the bounded operator $T_V$.

EDIT. Maybe it is worthwhile to mention the following generalization of 3) (which is inspired by a comment of Christian Remling below the question):

In fact, the same assertion remains true if we only assume that the resolvent is bounded outside a ball of sufficiently large radius. To see this, one argues similarly as in the proof of 3), but now Liouville's theorem only yields that the resolvent of $T_W$ is constant; so say $R(z,T_W) = R$ for a bounded operator $R$ on $W$. Now the resolvent identity implies that $R^2 = 0$. But $R^2$ is an injective operator, so we again obtain $W = 0$.