You are working with a less elegant definition of the lightface Borel hierarchy. Here is a more elegant one, and the relationship.
A set $A \subseteq \omega^\omega = \mathcal{N}$ is effectively open if it is the union of a computable enumerable sequence of basic open sets. This means there is a c.e. sequence $\langle \sigma_i : i \in \omega\rangle$ such that $ A = \{ f : (\exists i)[\sigma_i \subseteq f]\}$. A code for $A$ is a pair $(1, i)$ where $i$ is an index of the c.e. enumeration.
The pointclass $\Sigma^0_1$ consists of the effectively open sets. More generally:
The pointclass $\Pi^0_{n}$ consists of the complements of sets in $\Sigma^0_n$. The code for such a set is the same as the code for its complement.
The pointclass $\Sigma^0_{n+1}$ consists of all effective countable unions of sets in $\Pi^0_{n}$. This means that a set $A$ is $\Sigma^0_{n+1}$ if there is a computable sequence of codes of $\Pi^0_n$ sets such that $A$ is the union of these sets. The code for the $\Sigma^0_{n+1}$ set is a pair $(n+1, i)$ where $i$ is an index of the c.e. enumeration of codes of $\Pi^0_n$ sets.
This coding scheme can be continued to define the classes $\Sigma^0_\alpha$ and $\Pi^0_\alpha$ when $\alpha$ is a computable ordinal. But here we will stick to $\Sigma^0_n$ and $\Pi^0_n$ for $n$ in $\omega$.
The key fact is that these sets match up with the arithmetic hierarchy for formulas in second-order arithmetic:
Theorem. A subset of $\omega^\omega$ is lightface $\Sigma^0_n$ if and only if it is definable by a $\Sigma^0_n$ formula of second-order arithmetic (the formula will have one free function variable and no other free variables). The same holds for $\Pi^0_n$ sets, mutatis mutandis.
That answers your question, because if $R(m)$ is any arithmetically definable predicate then $\phi(f) \equiv (\exists n)R(f\upharpoonright n)$ is in the arithmetical hierarchy of formulas in second-order arithmetic, so the subset of $\omega^\omega$ that it defines is in the lightface Borel hierarchy.
The requirement that $R(f\upharpoonright n)$ be monotone is a consequence of the definition in the question using relations on $\omega$ instead of $\Sigma^0_1$ formulas of second-order arithmetic. There is a use principle for such formulas: if $f$ satisfies a $\Sigma^0_1$ formula $\phi(x)$ then there is a $k$ such that every $g$ with $g \upharpoonright k = f \upharpoonright k$ also satisfies $\phi$ (this is really just the claim that $\phi$ defines an open set).
The definition in the question avoids talking about formulas of second-order arithmetic, instead only talking about formulas of first-order arithmetic. This necessitates the assumption of monotonicity, because what the definition is really looking for is a definition of an effectively open set, although it avoids that terminology.
