Yes definitely.
You can generate equidistributed sequences on $\bf R$ in the same fashion that you would do on $[0,1) \simeq {\bf R}/{\bf Z}$: by using an ergodic transformation preserving the Lebesgue measure. Note that the Lebesgue measure on $\bf R$ has infinite mass and thus the situation is slightly more subtle on $\bf R$ than on $[0,1)$.
The study of transformations preserving a measure of infinite mass is the topic of infinite measure theory. The standard reference is the book of J. Aaronson, an introduction to infinite measure theory edited by the AMS. The first example of a conservative ergodic transformation from $\bf R$ to $\bf R$ preserving the Lebesgue measure given in the book (and perhaps the simplest example) is the Boole transformation $$x \mapsto x - {1\over x}.$$
Applying the Hopf ratio ergodic theorem, we deduce that for almost all $x\in \bf R$, the sequence given by $x_{n+1} = x_n -{1\over x_n}$, $x_0=x$ is equidistributed according to your definition (which is almost the standard one).
Another way to generate an equidistributed sequence is to use a random walk on $\bf R$. So let $X_1$,...$X_k$... be an iid sequence of real valued random variables defined on some space $(\Omega, {\cal F}, P)$ which are both integrable and with zero expectation. We need a "non-arithmetic" assumption on the law of the $X_i$ because we don't want that the variables take values in a discrete subgroup of $\bf R$. $P_X = {1\over 1+\alpha}(\delta_{-\alpha}+\alpha \delta_1)$ with $\alpha$ positive irrational is ok.
Theorem the sequence $(S_n(\omega))$ is equidistributed on $\bf R$ for almost all $\omega$.
I think that this result is stronger than the CLT. Maybe this can be deduced from the local CLT. Anyway, this follows from the conservativity and ergodicity of the lift of the shift on the skew-product $\Omega\times \bf R$, the same argument as above applies. In addition to the previous reference, there are probably a few probability books where this is discussed but I can't remember any from the top of my head.