With the convention in the reference implementation, the recurrence is
$$u_{j+1}:=c\oplus(u_j\lll1)\oplus(u_j\vee(u_j\ggg1))$$
where $c$ is the $n$-bit constant with all bits at $0$ except the rightmost (otherwise said $c=0^{n-1}\mathbin\|1$ ), $\oplus$ is bitwise XOR, $\vee$ is bitwise OR, $\lll$ and $\ggg$ are left and right rotations of the $n$-bit bistring before the operator by the number of bits after.
If we revert the direction of shifts, that merely mirrors the (circular) bit mapping, thus does not change the cycle structure.
Does "Rule 30 with bit toggle" have minimum cycle length of ${\cal O}(2^n)$ where $n$ is the length of the bit vectors?
No, since for odd $n$ there is a minimum cycle of length one. That fixed point has binary expression an alternation of $\frac{n+1}2\ 1$ and $\frac{n-1}2\ 0$ (in hex: 55…55 for $n\bmod 4=3$ or 15…55 for $n\bmod 4=1$). In the following we thus restrict to even $n$.
Exploring small $n$ shows no evidence towards the claim: the minimum cycle length is often $3$, and does not seem to skyrocket.
n length start
2 2 0x0
4 5 0x1
6 3 0x03
8 6 0x14
10 3 0x07C
12 5 0x42F
14 7 0x035D
16 33 0x2D34
18 3 0x03E43
20 27 0x00A28
22 3 0x07C87C
24 4 0x102040
26 14 0x0ABB343
28 5 0x2D1E5A3
30 3 0x03E43E43
32 7 0x1B3AFA05
34 3 0x07C87C87C
36 13 0x0217F5A73
For a random function, the expected size of the cycle starting from a random point is close to $\sqrt{\pi2^n/8}=\mathcal O(2^{n/2})$; see Flajolet&Odlyzko's Random mapping statistics. The smallest cycle is typically much smaller (though I don't know the expected distribution). Thus the claimed cycle length would be somewhat a surprise.
On the other hand, the function has very slow diffusion, thus is very far from a random function.
Here is a graph for $n=14$.
