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 proof works as follows: First we will construct a sequence $(a_n)_{n \in \omega} \in \omega^\omega$ and find an ordinal $\beta < \omega_1$ such that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, ,$$ where $\prod_{m \leq n} a_m \times \prod_{m > n} \omega =\{ f \in \omega^\omega \, \colon \, \forall m \leq n \,\, f(m) < a_m \}$. The idea behind this is that we can require a bound for finitely many values of an input function $f$ and still make $F(f)(\beta)$ arbitrarily large.

In the second step we construct a sequence $(b_n)_{n \in \omega} \in \omega^\omega$ such that $(b_n)_{n \in \omega} \geq (a_n)_{n \in \omega}$, and a sequence $(f_n)_{n \in \omega}$ such that $$\forall n \in \omega \, \colon \, f_n \in \omega^\omega \, \land \,(b_m)_{m \in \omega} \geq f_n \, \land \, F(f_n)(\beta) \geq n \,.$$ But this leads to a contradiction, since the monotonicity of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$.

**First step:**

We will construct $(a_n)_{n\in \omega}$ by induction. For the base case $n=0$ we claim that $\exists a_0 \in \omega \,\exists \alpha_0 \in \omega_1 \,  \forall A \in [\omega_1 \setminus\alpha_0]^{\leq \aleph_0} \colon F[\prod_{m \leq 0} a_0 \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A$. This means that already $F\restriction  \prod_{m \leq 0} a_0 \times \prod_{m>0} \omega$ 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 $F\restriction  \prod_{m \leq 0} n \times \prod_{m>0} \omega$ 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 noting that $\omega^\omega = \bigcup_{n \in \omega} \prod_{m \leq 0} n \times \prod_{m>0} \omega$, there cannot exist a $g \in \omega^\omega$ such that $F(g) \restriction B \geq f$.

Assume inductively that $a_0,..,a_n$ and increasing $\alpha_0,...,\alpha_n$ have already been defined, such that $\forall A \in [\omega_1 \setminus\alpha_n]^{\leq \aleph_0} \, \colon \, F[\prod_{m \leq n} a_m \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A$. With a similar argument as before, and again noting that $$\prod_{m \leq n} a_m \times \prod_{m>n} \omega = \bigcup_{k \in \omega} \prod_{m \leq n} a_m \times \prod_{m=n+1} k \times \prod_{m>n+1} \omega$$ we can show that $$\exists a_{n+1} \in \omega \, \exists \alpha_{n+1} \in \omega_1 \setminus \alpha_n \, \forall A \in [\omega_1 \setminus\alpha_{n+1}]^{\leq \aleph_0} \colon \, F[\prod_{m \leq n} a_m \times \prod_{m=n+1} a_{n+1} \times \prod_{m>n+1} \omega] \,\text{is cofinal in} \, \omega^A \, .$$ So we can find $a_{n+1}$ and $\alpha_{n+1} > \alpha_n$ as required.

Now let $\beta > \sup_{n \in \omega} \alpha_n$. It follows from our construction of the $(a_n)_{n \in \omega}$ that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, .$$

**Second step:**

It follows from our construction above that there exists $f_0 \in a_0 \times \omega^{\omega \setminus \{0\}}$ such that $F(f_0)(\beta) \geq 2020$ (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 from 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, $$a_0 \times... \times a_n \times \omega^{\omega \setminus \{0,...,n\}}:=\{f \in \omega^\omega \,  \colon \,  \forall m \leq n \, \, f(m)\leq a_m\}$$