JavaScript (ES6),  38  33 bytes

Saved 5 bytes thanks to @JoKing

Takes input as a string. Expects a space for an empty square and any other character for a rook.

s=>(s+s).trim().split` `.length-1

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Commented

s =>          // s = input, e.g. " R  RRR "

  (s + s)     // double -> " R  RRR  R  RRR "

  .trim()     // remove leading and trailing spaces -> "R  RRR  R  RRR"

  .split` `   // split on spaces -> [ 'R', '', 'RRR', '', 'R', '', 'RRR' ]

  .length - 1 // return the length - 1 -> 6

Python 2, 40 bytes

lambda s:len((s+s).strip().split(' '))-1

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Retina, 14 9 bytes

w`_+R|R_+

Try it online! Link includes test cases. Uses _ for empty space as it's the most pleasant non-regex character. Works by counting the number of substrings that correspond to a valid Rook move. A substring is a valid Rook move if it contains at least one _ plus a single R at either the beginning or the end.

Python 3, 30 29 bytes

lambda s:sum((s+s).strip())/9

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-1 byte thanks to @JoKing

The function takes a Python byte string as input. Each empty space is encoded as a tab and each rook is encoded as a byte b'x00' having value 0.

The computation is equivalent to lambda s:(s+s).strip().count(b't') while having a lower byte count.

Jelly, 6 bytes

t1;ḟẠS

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A monadic link taking a list of 0 for rook and 1 for space and returning an integer with the number of moves. The TIO link takes the pasted list of possible boards given in the question, converts to the right format and then outputs the calculated and correct answers.

Explanation

t1     | Trim 1s from end

  ;    | Concatenate to input

   ḟẠ  | Filter out 1s if all input were 1s, otherwise filter out 0s

     S | Sum

Python 2, 59 bytes

def f(r):S=map(len,r.split('R'));return sum(S)*2-S[0]-S[-1]

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Perl 6, 16 bytes

{+m:ex/s+R|Rs+/}

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A regex that matches all exhaustive instances of rooks followed by spaces, or spaces followed by a rook and returns the number of matches.

Retina, 23 15 bytes

Double the number of spaces between rooks, grep lines with at least one rook, then count the number of spaces.

R +R

$0$0

G`R

 

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Though the program uses spaces instead of periods, I added prefix code so that the test cases provided may be easily pasted in and used.

I was hoping I could use overlapping matches with (?<=R.*) | (?=.*R), but overlaps aren't quite that aggressive. It would need to count all possible ways a match could be obtained in order to return the correct result with that method.

Jelly, 6 bytes

ṣ0§+ƝS

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0 represent a rook, 1 represents an empty space.

Japt, 6 bytes

²x ¸ÊÉ

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Japt, 6 bytes

Spaces for spaces, any other character for rooks.

²x ¸ÊÉ

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Stax, 7 bytes

âE¶ƒÅ╒±

Run and debug it

Wolfram Language (Mathematica), 43 38 bytes

Count[Subsequences@#,{0,1..}|{1..,0}]&

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Port of Neil's Retina solution. Uses 1 for spaces and 0 for rooks.

05AB1E, 5 bytes

«ðÚð¢

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«       # concatenate the input with itself

 ðÚ     # remove leading and trailing spaces

   ð¢   # count spaces

C# (Visual C# Interactive Compiler), 28 bytes

x=>(x+x).Trim().Sum(d=>d)/32

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Japt, 5 bytes

²x èS

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²x èS        Implicit input string of U

²            U + U

 x           Remove trailing and leading whitespace

   èS        Number of spaces

Haskell, 68 58 54 bytes

(n#y)(a:x)|a<1=n+(0#1)x|m<-n+1=y+(m#y)x

(_#_)x=0

f=0#0

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C, 181 bytes

#include<stdio.h>

#include<string.h>

main(a,c,d,e){c=a=d=0;char*b=calloc(99,1);gets(b);do{if(b[a]=='R'){c+=e*(1+d);e=0;d=1;}if(b[a]=='.')e++;a++;}while(b[a]);c+=e;printf("%i",c);}