The previous answer contained a major error/misconception, and I apologize for the dealy in correcting it. The answer to the question is "yes" locally in the Zariski topology but "no" globally.
Comparison of Milnor K-cohomology and motivic cohomology via edge maps:
Motivic cohomology has Zariski descent and hence there is a
descent/hypercohomology/Brown-Gersten-Quillen/coniveau spectral sequence computing the motivic cohomology of a smooth scheme $X$. On the $E_1$-page, the Gersten resolution of the sheaves $\mathcal{H}^q(\mathbb{Z}(n))$ (Zariski sheafification of motivic cohomology) appears. The entry $E_1^{p,q}$ is $\bigoplus_{z\in X^{(p)}}{\rm H}^{q-p}(\kappa(z),\mathbb{Z}(n-p))$ and the differential $d_1:E_1^{p,q}\to E_1^{p+1,q}$ is given by the appropriate residue maps.
The $E_2$-spectral sequence has the form
$$
E^{p,q}_2={\rm H}^p(X,\mathcal{H}^q(\mathbb{Z}(n)))\Rightarrow {\rm H}^{p+q}(X,\mathbb{Z}(n))
$$
and the differentials go $d_s^{p,q}:E^{p,q}_s\to E^{p+s,q-s+1}_s$.
This spectral sequence can then be used to relate Milnor K-cohomology to motivic cohomology. The relevant thing to note is the identification
$$
\tau_{\geq s}\mathbb{Z}(s)_X\to\mathbf{K}^{\rm M}_{s,X}[-s]
$$
which can be found e.g. in Section 3.1, Corollary 3.2 of
- K. Rülling and S. Saito. Higher Chow groups with modulus and relative Milnor K-theory. arXiv:1504.02669v2.
This statement basically says that $\mathcal{H}^q(\mathbb{Z}(n))=0$ for $q>n$ and $\mathcal{H}^n(\mathbb{Z}(n))=\mathbf{K}^{\rm M}_n$, where $\mathcal{H}^q$ denotes the motivic cohomology sheaves. (It also says that the answer is yes locally in the Zariski topology.)
From this vanishing statement, we get edge maps ${\rm H}^p(X,\mathbb{Z}(n))\to {\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$ because in the line $q=n$ there are no incoming differentials and so $E^{p,n}_\infty$ is a subgroup of ${\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$. Since $E^{p,n}_\infty$ is the last subquotient of the filtration of ${\rm H}^p(X,\mathbb{Z}(n))$ this provides the above edge maps. These would be the natural comparison maps (and they are also mentioned in Thi's answer.)
Comparison isomorphisms for $r=s$ and $r=s-1$:
To get the Bloch formula and the identification of ${\rm H}^{2n-1}(X,\mathbb{Z}(n))\cong {\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$, we need to show that there can be no outgoing differentials either. In the $E_2$-page, the relevant differentials for Bloch's formula are $d_s^{p,q}:{\rm H}^n(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$. The latter group is trivial, since on the $E_1$-page, the coefficients would be $\mathcal{H}^{1-2s}(\mathbb{Z}(-s))$ and the sheaves $\mathbb{Z}(-s)$ are trivial for $s>0$. The other relevant differentials are $d_s^{p,q}:{\rm H}^{n-1}(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s-1}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$ with the coefficients in the $E_1$-page given by ${\rm H}^{2-2s}(\mathbb{Z}(1-s))$. For $s>1$, this is again trivial. In particular, all the outgoing differentials are trivial, so that $E^{n,n}_\infty={\rm H}^n(X,\mathbf{K}^{\rm M}_n)$ and $E^{n-1,n}_\infty={\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$. This also implies that the only $E^{n+s,n-s}_\infty$ contributing to ${\rm H}^{2n}(X,\mathbb{Z}(n)))$ is ${\rm H}^n(X,\mathbf{K}^{\rm M}_n)$, yielding Bloch's formula. To get ${\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)\cong {\rm H}^{2n-1}(X,\mathbb{Z}(n))$ we need to show that ${\rm H}^{n+s-1}(X,\mathcal{H}^{n-s}(\mathbb{Z}(n)))=0$. The relevant coefficients in the $E_1$-page are $\mathcal{H}^{1-2s}(\mathbb{Z}(1-s))$. For $s=1$, we get $\mathcal{H}^{-1}(\mathbb{Z}(0))$ which is trivial, all the other terms are again trivial because $\mathbb{Z}(1-s)=0$ for $s>1$.
Failure of global comparison:
The natural comparison isomorphism fails to be an isomorphism in general. The simplest counterexample (pointed out by Jens Hornbostel) is actually $\mathbb{P}^1$. In this case, we have
$$
{\rm H}^r(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong \left\{\begin{array}{ll} {\rm K}^{\rm M}_s(F) & r=0\\
{\rm K}^{\rm M}_{s-1}(F) & r=1\end{array}\right.
$$
where $F$ denotes the base field. On the other hand, the projective bundle formula for motivic cohomology implies
$$
{\rm H}^{r+s}(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm H}^{r+s}(F,\mathbb{Z}(s))\oplus {\rm H}^{r+s-2}(F,\mathbb{Z}(s-1)).
$$
For $r=0$, we have ${\rm H}^0(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong {\rm K}^{\rm M}_s(F)$ and ${\rm H}^s(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm K}^{\rm M}_s(F)\oplus {\rm H}^{s-2}(F,\mathbb{Z}(s-1))$, so Milnor K-cohomology and motivic cohomology differ whenever ${\rm H}^{s-2}(F,\mathbb{Z}(s-1))\neq 0$. A particular instance where that happens is $s=3$ in which case ${\rm H}^1(F,\mathbb{Z}(2))\cong {\rm K}^{\rm ind}_3(F)$.
In the formulation of the descent spectral sequence, the spectral sequence is contained in the two columns for $p=0,1$ and degenerates at the $E_2$-term. The failure of the global comparison follows since $\mathcal{H}^2(\mathbb{Z}(3))$ has nontrivial first cohomology over $\mathbb{P}^1$, given by ${\rm H}^1(F,\mathbb{Z}(2))$.
Best Answer
At the time when I constructed those maps, the motivic spectral sequence was not known to exist. I gave a conjectural construction for that spectral sequence in an American Journal of Math paper called "Relative Cycles and Algebraic K-Theory", but I still don't know whether the gaps in that construction can be (or have been) filled.
I stopped working on this long enough ago that I could be remembering wrong, but my best recollection is that if the conjectural construction in RCAKT can be made to work, and if the spectral sequence it yields is the same as Friedlander-Suslin's (which it certainly ought to be), then you can piece together results in those two papers to show that the maps in question are compatible with the differentials. (You might also need some results from my paper "Some Filtrations on Higher K-Theory and Related Invariants".)
There might also be a much easier argument, directly using the Friedlander-Suslin construction, which would save you from mucking around in those other papers.
I apologize for many things, including my hazy memory, my failure to state this result explicitly in the original papers, and the fact that I'm not up to date on the current state of the art.