$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\cO{\mathcal{O}}\def\FF{\mathbb{F}}$Does this help?
$K(X) \otimes \mathbb{Q}$ is a graded ring in the sense that it is isomorphic (by the Chern character map) to $A(X) \otimes \mathbb{Q}$, which is graded. I would perhaps prefer to say that it is "gradeable", since the grading isn't very obvious in terms of $K$-theory. The most $K$-theoretic way I know to describe it is that the Adams operators $\psi^k$ act by $k^j$ on the $j$-th graded piece. The corresponding descending filtration is the filtration by codimension.
For example, let $L$ be a line bundle with $c_1(L) = D$. Then $ch(L) = e^D$. So $ch(\log L)
=D$ and $\log [L]$, defined as the class $([L]-1) - ([L]-1)^2/2 + ([L]-1)^3/3 - \cdots$ in $K(X)$, is pure of degree $1$. Indeed, $\psi^k [L] = [L^k]$, so $\psi^k \log [L] = \log [L^k] = k \log [L]$. Let's abbreviate $\log [L]$ by $z$. The structure sheaf of the vanishing locus of a section of $L$ has $K$-class $1-L^{-1} = 1-e^z = z-z^2/2+z^3/6-z^4/24+\cdots$. So this class is in the filtered part that has degree $\geq 1$, but is not of pure degree.
A natural question, to which I don't know the answer, is whether the integer $K$-ring has a natural grading $K(X) \cong \bigoplus K_i(X)$, which turns into this grading when we tensor by $\mathbb{Q}$.
I now have some concrete examples of algebraic varieties where $K(M, \mathbb{Z})$ is not isomorphic to $A(M, \mathbb{Z})$, and where $K(M, \mathbb{Z})$ does not support a grading with the correct Betti numbers.
The first example, due to Matt Larson, is to take $Q$ a quadratic $3$-fold in $\mathbb{P}^4$. Then $Q$ is a homogenous space for $SO(5)$, so it has a cellular stratification, as follows: Consider a chain $P \subset L \subset H \subset Q$, where $P$ is a point, $L$ is an isotropic line, $H$ is a singular hyperplane section containing $L$, and $Q$ is the whole space. Then $L\setminus P \cong \mathbb{A}^1$, $H \setminus L \cong \mathbb{A}^2$ and $Q \setminus H \cong \mathbb{A}^3$. The singular hyperplane section $H$ is rationally equivalent to a smooth hyperplane section $H'$ through $L$, in which case $H'$ is a quadratic surface and $L$ is one of the two rulings of $H'$; I'll prefer $H'$ for multipication computations, but $Q \setminus H'$ isn't $\mathbb{A}^3$, so I need $H$ to get the cellular stratification.
Thus, the Chow groups of $Q$ are $(\ZZ [Q], \ZZ [H], \ZZ [L], \ZZ [P])$. The non-obvious products are
$$[H]^2 = 2[L],\ [H][L] = [P].$$
To see the former relation, the normal bundle to $H'$ is $\mathcal{O}(1,1)$, so the self-intersection of $H'$ is the sum of a line from one ruling and a line from the other ruling, both of which are rationally equivalent to $L$ within $Q$.
$K^0(Q)$ also has $\ZZ$-basis the structure sheaves of $Q$, $H$, $L$ and $P$. This time, the products are
$$[\cO_H]^2 = 2[\cO_L]-[\cO_P],\ [\cO_H][\cO_L] = [\cO_P].$$
To see the former relation, note that the self-intersection of $\cO_{H'}$ is the sum of one line from each ruling, minus the point where they intersect.
If these rings were isomorphic, their tensor products with $\mathbb{F}_2$ would be isomorphic. A useful invariant of an $\mathbb{F}_2$ algebra is the dimension of the vector space spanned by the squares of all elements. In Chow, we have $[H]^2 \equiv [L]^2 \equiv [P]^2 \equiv 0 \bmod 2$, so this vector space is $1$-dimensiona with basis $1$; in $K$-theory, we have $[\cO_H]^2 \equiv [\cO_L] \bmod 2$ and $[\cO_L]^2 \equiv [\cO_P]^2 \equiv 0 \bmod 2$, so this vector space is $2$-dimensional with basis $1$, $[\cO_L]$. (Because squaring is a linear map modulo $2$, we only need to square the generators.)
Let's see further that the $K$-theory ring can't be graded as $K_0(Q) = \ZZ \oplus \ZZ \gamma_1 \oplus \ZZ \gamma_2 \oplus \ZZ \gamma_3$, with $\gamma_i$ in degree $i$. Degree considerations show that $\gamma_2$ and $\gamma_3$ square to $0$, so they must lie in $\ZZ [\cO_L] + \ZZ [\cO_P]$, and thus must span it. Since $\gamma_1$ is nilpotent, it must be of the form $a [\cO_H] + b[\cO_L] + c[\cO_P]$; since $\ZZ \gamma_1 \oplus \ZZ \gamma_2 \oplus \ZZ \gamma_3$ is the maximal ideal, we must have $a = \pm 1$ and, WLOG, we can take $a=1$. Then $\gamma_1^2$ should be $m \gamma_2$ for some integer $m$. But $\gamma_1^2 = ([\cO_H] + b[\cO_L] + c[\cO_P])^2 = 2 [\cO_L] + (2b-1) [\cO_P]$ and $GCD(2, 2b-1) = 1$ so we must have $m=1$ and $\gamma_2 = 2 [\cO_L] + (2b-1) [\cO_P]$. Similarly, $\gamma_3$ must be $\gamma_1^3/2 = [\cO_P]$. But $([\cO_H] + b[\cO_L] + c[\cO_P], 2 [\cO_L] + (2b-1) [\cO_P], [\cO_P])$ only span an index $2$ sublattice in $K^0$, a contradiction.
I have asked various people for toric counterexamples, and now I have one. I'll do this more quickly. Let $C$ be the toric variety whose fan is combinatorially the normal fan to the cube, but with rays in directions
$$(1,0,0), (0,1,0), (0,0,1), (-1,0,0), (1,-1,0), (0,2,-1).$$
This is a so-called "Bott tower".
This space is a tower of $\mathbb{P}^1$ bundles, so I can compute $K$-theory and cohomology/Chow using the projective bundle theorem. (Cohomology) ($K$-theory).
For cohomology/Chow, I get
$$H^{\ast}(C) = A^{\ast}(C) = \ZZ[X,Y,Z]/\langle X^2, Y(Y+X), Z(Z+2Y) \rangle$$
In other words,
$$X^2=0, Y^2 = -XY,\ Z^2 = -2YZ.$$
Note that the square free monomials in $X$, $Y$, $Z$ are a basis.
In $K$-theory, I get
$$K^{0}(C) = \ZZ[u,v,w]/\langle (u-1)^2,\ (v-1)(vu-1),\ (w-1)(wv^2-1) \rangle.$$
Put $u = 1+x$, $v = 1+y$ and $w = 1+z$. Since $u$, $v$, $w$ are classes of line bundles, $x$, $y$ and $z$ are in codimension $1$. We get the relations
$$x^2 = 0,\ y(y+x+xy)=0,\ z(z+2y+y^2+2yz+y^2z)=0.$$
So
$$x^2=0$$
$$y^2 = -xy-xy^2 = -xy+x(xy+xy^2) = -xy.$$
$$z^2 = -2yz - y^2 z - 2y z^2 = -2yz + xyz - 2y(-2yz +\cdots) = -2yz + xyz +4 y^2 z = -2yz - 3xyz.$$
Here the "$+\cdots$" are terms that I know will die at the next step for dimension reasons. To summarize
$$x^2=0,\ y^2 = -xy,\ z^2 = -2yz-3xyz.$$
We can see that the Chow relations are the associated graded of the $K$-theory relations.
However, we can also see that the rings are not isomorphic. Tensor
with $\FF_2 = \ZZ/2 \ZZ$ to get
$$\FF_2[X,Y,Z]/(X^2=Z^2=0,\ Y^2 = YZ)$$
and
$$\FF_2[x,y,z]/(x^2=0, y^2=yz, z^2 = xyz)$$
respectively. The two rings are not isomorphic, because there squares form a $2$-dimensional $\FF_2$ vector space in the first case (basis $\{ 1, YZ \}$) and $3$-dimensional vector space in the second (basis $\{ 1,yz, xyz \}$).
