I have seen somewhere that if a differential graded algebra $A$ is connective (homologically graded), then the Grothendieck group $K_{0}(A)=K_{0}(H_{0}(A))$.
Suppose that $A$ is a differential graded algebra such that $A$ is connective (cohomologically graded) then the Grothendieck group $K_{0}(A)=K_{0}(H^{0}(A))$ ?
If I'm not wrong a differential graded algebra such that $A$ is connective in cohomological grading is called coconnective.
My previous answer was wrong, but this recent preprint examines the question in detail (for the whole $K$-theory spectrum).
A counterexample to your specific question is given as Example 5.15, where the authors find a coconnective algebra $R$ with $K(R)\simeq K(\mathbb A^n_k)\oplus K(k)$ (for some field $k$), and where the map $\pi_0R\to R$ induces the inclusion of the left summand. In particular, on $K_0$, they differ by $K_0(k)\cong \mathbb Z$.
The algebra in question is the homotopy pullback of $\Gamma(\mathbb A^n)\times_{\Gamma(\mathbb A^n \setminus \{0\})} \Gamma(\mathbb A^n)$ where $\Gamma$ means global sections.
The paper does give good conditions under which the map $\pi_0R\to R$ induces an equivalence on the full $K$-theory spectrum, see their Theorem 1.1.
Note that in algebra one usually writes $H^{-n}$ instead of $H_n,$ so connective and coconnective algebras live in the same bigger category of (cohomologically graded) DGA's. (There is also a topological convention of writing $\pi_{-n}$ instead of $H^n$). If $A$ is connective is has cohomology in negative degrees, and there is a map $A\to H^0(A).$ If $A$ is coconnective, then there is a map $H^0(A)\to A.$
Now $K$ theory is properly an invariant of a(n $\infty$-) category, not an algebra. It is never interesting to take the category of all $A$-modules (whose $K$ theory is trivial by the Eilenberg swindle), so you always want to impose some finiteness condition. Usually the $K$ theory of $A$ is defined as the $K$ theory of the category of perfect modules. In this case, given a map $f:A\to B$ of DGA's, the functor of restriction $B\mathrm{mod}\to A\mathrm{mod}$ does not (in general) preserve perfect objects, so the only natural map on $K$ theory is the one associated to the induction functor, $M\mapsto M\otimes_A B$ from $A$-modules to $B$-modules. Induction from $H^0(A)\to A$ in the coconnective case does not induce an isomorphism on $K$ theory in the very simplest nontrivial coconnective dga, $k\oplus k[1]$ (square zero extention), where the induced map on $K^0$ is trivial (any perfect module over $k$ has equal total dimension of even and odd cohomology). In the connective case, the map on $K$ theory goes from $K(A)\to K(H^0(A))$. I can't think of an obvious counterexample but I actually don't think think this functor induces an isomorphism on $K^0$ in general despite what you heard; I may be wrong.
However there are other categories of modules that one can study. In particular, in the coconnective case it is interesting to study the category of modules which are perfect over $H^0(A)$ (or, more generally, whose cohomology groups are finitely generated over $H^0(A)$); this is related to the coherent $K$ theory of a scheme.
Now in these cases, the restriction functor along $H^0(A)\to A$ gives a valid map on $K$ theory, and moreover induces an isomorphism on $K^0$ as well as on all higher $K$ groups.
