It is known that, for any plain rectangular sliding puzzle (larger than 2x2), any parity-conforming configuration is reachable from the solved state, and all of those with wrong parity are unreachable. "Parity-conforming" means that the parity of the entire board, plus the Manhattan distance of the hole from its solved position, must be even. Relevant info can be found on Wikipedia and MathWorld.
The CORONA board has two distinctions from a plain 2x4 board: there are two O's, and they're stuck together. I observed that the vertical OO in the middle can change its own parity, but only by changing the location of the hole (relative to the blocking vertical OO) at the same time. When OO is somewhere at the border (either horizontal or vertical), it looks like the pair can have only one orientation.
In order to swap two O's in place, we need to move the piece like a T-junction.
?? O1 ?? ?? => ?? ?? ?? ?? => ?? ?? ?? ?? => ?? O2 ?? ??
?? O2 ?? ?? O2 O1 ?? ?? ?? O2 O1 ?? ?? O1 ?? ??
However, note that the first move requires a hole to be present on the left side, and the third requires a hole on the right side. A hole cannot move across the vertical piece without moving the vertical piece first.
Also, assume the first state has the hole on its left. Regardless of how you move the holes around, moving the OO will result in one of the following two states:
O1 O2 ?? ?? or ?? ?? ?? ??
?? ?? ?? ?? O2 O1 ?? ??
and moving it from the last state will result in one of the following:
?? O1 O2 ?? or ?? ?? ?? ??
?? ?? ?? ?? ?? O2 O1 ??
all of which orient O2 on the clockwise direction from O1. There is no way to flip the orientation of OO on the border.
To summarize all of the above, given the position of the OO and the hole, the orientation of OO is fixed. This means that the parity argument applies even though two O's are present, declaring all the odd-parity states unreachable.
I think proving that all the even-parity states are reachable is easy: just go through a state where OO is placed vertically on the left side, and freely move the remaining 2x3 portion of the board.
Now to the actual task. If non-words are allowed,
I can get CRAOON in 16 steps:
1 4 1 3
C O R x => C . R x => C R x A => C R x A =>
. O N A O O N A O O . N . O O N
1 4 1 1
R x . A => R x O A => C R O A => C R . A => C R A .
C O O N C . O N x . O N x O O N x O O N
For an actual word,
I'm getting CORONA again, in 19 moves (starting with the state after 15 moves):
C R . A => C O R A => C O R .
x O O N x O . N x O N A
I guess the message from this puzzle is that
CORONA is persistent.
jafe suggested the word (then confirmed by OP)
which turned out to be possible but apparently too far away (slightly improved after edit):
32 30 28 moves.
3 3 3 3
C O R x => O C R x => O . C x => O N . x =>
. O N A O . N A O N R A O R C A
O N x A => O x R A => R A C .
O R . C O . N C x O O N
1 5 6 4
C O R x => C . R x => C R x A => R x A N =>
. O N A O O N A . O O N C . O O
5 6 3
R A N O => A C N O => R A C N => R A C .
C x . O R . x O . x O O x O O N
2 2 2 4
C O R x => O O R x => O . R x => O R N x =>
. O N A C . N A O C N A O C . A
4 3 6 9
O R A N => O A C N => O A . C => O x R C => R A C .
O C . x O R . x O R x N O . A N x O O N
According to the result of anagram search on Qat, the two words already presented are the only reachable words.
CAROON has wrong parity (because RACOON is reachable and CAROON is RACOON with R and C swapped), and ORACON would tear apart the two O's.