The answer is no, if you require that the monotone function $F \, \colon \, \omega^\omega \rightarrow \omega^{\omega_1}$ is total. I will use fairly standard notation, i.e. $f,g \in \omega^\omega$, $\alpha, \beta \in \omega_1$ and $m.n \in \omega$. The main idea is to use many diagonalization arguments.
Claim: $\exists a_0 \in \omega \,\exists \alpha_0 \in \omega_1 \, \forall A \in [\omega_1 \setminus\alpha_0]^{\leq \aleph_0} \colon F[a_0 \times \omega^{\omega \setminus \{0\}}] \,\text{is cofinal in} \, \omega^A$. This means that already $F\restriction a_0 \times \omega^{\omega \setminus \{0\}}$ dominates $\omega^A$ for every countable $A \subseteq \omega_1$ above $\alpha_0$.
Towards a contradiction assume that the claim is wrong. Therefore, we can construct a sequence $(A_n)_{n \in \omega}$ such that $A_n \in [\omega_1]^{\leq \aleph_0}$ and $\sup A_n < \min A_{n+1}$, and a sequence of functions $(f_n)_{n \in \omega}$ such that $f_n \in \omega^{A_n}$ and $F\restriction n \times \omega^{\omega \setminus \{0\}}$ does not dominate $f_n$. But if we define $B:= \bigcup_{n \in \omega} A_n \in [\omega_1]^{\leq \aleph_0}$ and $f:= \bigcup_{n \in \omega} f_n \in \omega^B$, we reach a contradiction, since there is no $g \in \omega^\omega$ such that $F(g) \restriction B \geq f$.
By induction we can now construct a sequence $(a_n)_{n \in \omega}$, $a_n \in \omega$, and an increasing sequence $(\alpha_n)_{n \in \omega}$, $\alpha_n \in \omega_1$ such that $\forall n \in \omega \, \colon$ `$F[a_0 \times... \times a_n \times \omega^{\omega \setminus \{0,...,n\}}]$ dominates $\omega^A$ for every countable $A \subseteq \omega_1$ above $\alpha_n$'. In the induction step simply repeat the proof of the claim with $a_0 \times... \times a_n \times \omega^{\omega \setminus \{0,...,n\}}$ instead of $\omega^\omega$, $\omega_1 \setminus \alpha_n$ instead of $\omega_1$ and noticing that $$a_0 \times... \times a_n \times \omega^{\omega \setminus \{0,...,n\}}= \bigcup_{m \in \omega} a_0 \times... \times a_n \times m \times \omega^{\omega \setminus \{0,...,n+1\}}$$
Now let $\beta > \sup_{n \in \omega} \alpha_n$. It follows from our construction above that there exists $f_0 \in a_0 \times \omega^{\omega \setminus \{0\}}$ such that $F(f)(\beta) \geq 2020$ (as $\geq 0$ would be trivial). Set $b_0:= a_0$. Assume inductively that $b_0,...,b_n$ and $f_0,...,f_n$ have already been constructed such that $$\forall m \leq n \, \colon \, a_m \leq b_m \, \land \, f_m \in b_0 \times... \times b_n \times \omega^{\omega \setminus \{0,...,n\}} \land \, F(f_m)(\beta) \geq m$$ Again by using our construction above, we can now find $f_{n+1} \in b_0 \times... \times b_n \times \omega^{\omega \setminus \{0,...,n\}}$ such that $F(f_{n+1})(\beta) \geq n+1$ and set $b_{n+1}:= \max( \max_{m \leq n+1} f_m(n+1), a_{n+1})$.
But now we have reached a contradiction, since the monotonicity (and totality) of $F$ implies that $\forall m \in \omega \, \colon \, F((b_n)_{n \in \omega})(\beta) \geq m$.
PS: I have just now realized that I slightly misuse notation. Of course, $$a_0 \times... \times a_n \times \omega^{\omega \setminus \{0,...,n\}}:=\{f \in \omega^\omega \, \colon \, \forall m \leq n \, \, f(m)\leq a_m\}$$