Let me first write what happens for classical cobordism. You are basically asking whether the map $\operatorname{MGL}\to H\mathbb{Z}$ factors through the projection $\operatorname{ku}\to H\mathbb{Z}$. But this is clear, since all spectra in sight are connective and we have an equivalence $$\operatorname{Map}(E,H\mathbb{Z})\cong \operatorname{Map}(\pi_0E,\mathbb{Z})$$ for every connective spectrum $E$ given by the fact that $\pi_0$ is the left adjoint of the inclusion of discrete spectra (i.e. abelian groups) into connective spectra. This adjunction by the way works both in spectra and abelian groups and $E_\infty$-ring spectra and commutative rings, in which case it's clear that the right hand side is just a point both for $\operatorname{MGL}$ and $\operatorname{ku}$.

Something similar happens in motivic cohomology. Here the statement is that the map $\operatorname{MGL}\to H\mathbb{Z}$ factors uniquely through the orientation $\operatorname{MGL}\to\operatorname{kgl}$ (here with $\operatorname{kgl}$ we mean the very effective cover of the motivic spectrum $\operatorname{KGL}$). Instead of taking homotopy groups the correct thing to do is taking slices. Indeed the both map $\operatorname{MGL}\to H\mathbb{Z}$ and $\operatorname{kgl}\to H\mathbb{Z}$ exhibit $H\mathbb{Z}$ as the zero slice of the respective spectrum and $\operatorname{MGL}\to\operatorname{kgl}$ induces an equivalence on 0-slices.

Let me first write what happens for classical cobordism. You are basically asking whether the map $\operatorname{MU}\to H\mathbb{Z}$ factors through the projection $\operatorname{ku}\to H\mathbb{Z}$. But this is clear, since all spectra in sight are connective and we have an equivalence $$\operatorname{Map}(E,H\mathbb{Z})\cong \operatorname{Map}(\pi_0E,\mathbb{Z})$$ for every connective spectrum $E$ given by the fact that $\pi_0$ is the left adjoint of the inclusion of discrete spectra (i.e. abelian groups) into connective spectra. This adjunction by the way works both in spectra and abelian groups and $E_\infty$-ring spectra and commutative rings, in which case it's clear that the right hand side is just a point both for $\operatorname{MU}$ and $\operatorname{ku}$.

Something similar happens in motivic cohomology. Here the statement is that the map $\operatorname{MGL}\to H\mathbb{Z}$ factors uniquely through the orientation $\operatorname{MGL}\to\operatorname{kgl}$ (here with $\operatorname{kgl}$ we mean the very effective cover of the motivic spectrum $\operatorname{KGL}$). Instead of taking homotopy groups the correct thing to do is taking slices. Indeed the both map $\operatorname{MGL}\to H\mathbb{Z}$ and $\operatorname{kgl}\to H\mathbb{Z}$ exhibit $H\mathbb{Z}$ as the zero slice of the respective spectrum and $\operatorname{MGL}\to\operatorname{kgl}$ induces an equivalence on 0-slices.