The usual $H$-convergence is defined for operators of the following form (for the sake of simplicity, restrict ourselves with the one-dimensional case): $$ \frac{d}{d x}\left[A_\varepsilon(x) \frac{d w_\varepsilon}{dx}\right] = f, $$ where $A_\varepsilon$ is usually assumed to be an $L^\infty$ function (with some additional constraints), $w_\varepsilon \in H^1$ and $f \in H^{-1}$ is a distribution. Then, $A_\varepsilon$ is said to be $H$-convergent to an $A_0 \in L^\infty$, if and only if $$ w_\varepsilon \to w_0 ~~ as ~~ \varepsilon \to 0 ~~ weakly ~~ in ~~ H^1, $$ $$ A_\varepsilon \frac{d w_\varepsilon}{d x} \to A_0 \frac{d w_0}{d x} ~~ as ~~ \varepsilon \to 0 ~~ weakly ~~ in ~~ L^2. $$ Here $w_0$ is the solution of $$ \frac{d}{d x}\left[A_0(x) \frac{d w_0}{dx}\right] = f. $$
I am interested if this concept can be generalized to be valid for $A_\varepsilon$ which converges to $A_0$ in the sense of distributions, i.e., for any compactly supported smooth function $\varphi$, $$ \int_{-\infty}^\infty A_\varepsilon(x) \varphi(x) dx \to \int_{-\infty}^\infty A_0(x) \varphi(x) dx ~~ as ~~ \varepsilon \to 0? $$
What follows is an expansion of the comments made above earlier. In general, there is little to no relevance between homogenization of the equation $$ \tag{1} - \frac{d}{dx} \left(A_\varepsilon (x) \frac{d w_\varepsilon(x) }{dx} \right) = f $$ and the actual convergence between $A_\varepsilon$ to $A_0$ (in some metric, for instance).
The model case, to keep in mind is the case of periodic homogenization, when one assumes that $A:\mathbb{R} \to \mathbb{R}$ is $C^\infty$ and $1$-periodic, i.e. $A(x + h) = A(x)$ for all $x\in \mathbb{R}$ and any $h \in \mathbb{Z}$, and then defines $A_\varepsilon (x) = A(x/ \varepsilon)$.
Assuming $A$ to be elliptic, and setting the problem $(1)$ in some interval $(a,b) \subset \mathbb{R}$ with $f\in L^1(a, b)$ and $w_\varepsilon \in H^1_0(a,b)$ ($0$-boundary data) it is a well-known fact (see any book on homogenization of PDEs, for instance the classics) that $$ A_0 = \frac{1}{ \int\limits_0^1 \frac{dx}{A(x)} }. $$ Notice, that we are in 1d and ellipticity condition on $A$ forces $A>0$.
So, we have that $A_\varepsilon$ converges to $\int\limits_0^1 A(x) dx$ weak* in $L^\infty$ (this is not hard to see, using for example the Fourier series expansion of $A$), however the actual homogenized coefficient $A_0$ is different from the weak limit of $A_\varepsilon$.
The point being made in the model case discussed above, is that one should think about the convergence business related to $A_\varepsilon$ from the perspective of homogenization: you do not aim to find $A_0$ to which $A_\varepsilon$ actually converges, but rather you find $A_0$ from the condition that the entire problem $(1)$ "converges" to the one with coefficient $A_0$. That is, there is a convergence of the solutions of $\varepsilon$-problem to the one with homogenized coefficients. In particular the $H$-convergence you mentioned, refers to the homogenization of the problem $(1)$ and does not necessarily imply direct convergence of coefficients $A_\varepsilon$ to $A_0$ in any standard metric.