I doubt it. Let's first dissect your Cauchy-Crofton formula. Up to minor technical assumptions, what we have is that
$$\mbox{length}(\gamma) = \int_{\mathbb{R}^2} \delta_\gamma(x) d^2x$$
where $\delta_\gamma(x)d^2x$ represents the measure concentrated on $\gamma$. Locally you can think of its as the pull back of the Dirac delta via the charaterization function of $\gamma$ as a submanifold.

Next we do a change of coordinates. Observe that (not being too precise here) $\mathbb{R}^2 = TS^1 \times \mathbb{R} / S^1$ (basically you can rotate a standard coordinate system; in other words, the set of all lines on the plane can be identified with $TS^1$, and we take its Cartesian product with $\mathbb{R}$ to measure along each of the lines). So let $\pi$ be the canonical projection map from $TS^1\times \mathbb{R}$ to $\mathbb{R}^2$. Then your integral can be re-written as
$$ \mbox{length}(\gamma) = \frac{1}{|S^1|}\int_{TS^1\times \mathbb{R}} \pi^*\delta_\gamma(y) dy$$
where $\pi^*\delta_\gamma(y) dy$ is the pull back measure. Now you integrate out the fibers $\mathbb{R}$ first and by simple geometry you see that
$$ \int_{TS^1\times\mathbb{R}}\pi^*\delta_\gamma(y) dy = \int_{TS^1} n(s) ds $$
where $n(s)$ is the number of times the line indexed by the coordinate $s$ intersects $\gamma$. So up to some normalization constants we recover the Cauchy-Crofton formula.

Now, if you want to get an integral of the curvature $k$ along the curve, you can write it as
$$ \mbox{curvature integral} = \int_{\mathbb{R}^2} |k(x)|^p \delta_\gamma(x) d^2x$$
the procedure of lifting to the space of lines is no problem, so you can again get an integral over $TS^1\times \mathbb{R}$. The problem is that I can't see any obvious way of integrating out the additional factor of $\mathbb{R}$. In the Cauchy-Crofton case, each time the line hits your curve it picks up a unit bundle of mass. In the curvature case, you pick up some number which depends on the curvature at the point of intersection.

Using circles won't help either. You can consider a foliation (or almost a foliation) of $\mathbb{R}^2$ by some family of curves, each of which can be identified with some curve $m$. Then you can do the same thing as above: let $M$ be the parameter space of this foliation (the foliation can be moved under rigid motion transformations to get a new foliation) then $\mathbb{R}^2$ can be identified as $M\times m / G$ by $G$ being some subgroup of the Euclidean symmetries. So formally your integral can be re-written as
$$ \frac{1}{|G|}\int_{M\times m} \mbox{something} dy $$
but to simplify down to just an integral over the parameter space $M$, you need to integrate out along the fibre $m$, and to get a simple expression at the end you almost certainly need that the family of curves $m$ in your foliation must be very special. Unless $m$ is adapted to $\gamma$ such that the integral along $m$ of $|k|^p\delta_\gamma$ can be easily evaluated, you have no hope of arriving at a simple integral expression.