Python 2/3, 236 bytes

236 byte version:

n=input();r=len;u=[("",[0]*(n//4))]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1;
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

This is fairly fast, for a brute-force-ish method, only taking a few seconds for n=223, but much longer for n>=224.

Explanation: Keep track of a list of string-list pairs (s,u), where the list u is such that u[i] is the current position after following every ith step in the string. For each string in the list, try to add "L" or "R", then change the values in the list that intersect. (i.e. if the resulting string has length 10, add or subtract 1 from positions 1,2,5 and 10, according to the directions you moved). If you exceed 3 or -3 throw the new pair away, otherwise keep it in the list. The longest strings are kept at the end. Once you have a string of length n, return it.

Python 2, 236 bytes

n=input();r=len;u=[("",[0]*(n//4))]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1;
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

This is fairly fast, for a brute-force-ish method, only taking a few seconds for n=223, but much longer for n>=224.

Explanation: Keep track of a list of string-list pairs (s,u), where the list u is such that u[i] is the current position after following every ith step in the string. For each string in the list, try to add "L" or "R", then change the values in the list that intersect. (i.e. if the resulting string has length 10, add or subtract 1 from positions 1,2,5 and 10, according to the directions you moved). If you exceed 3 or -3 throw the new pair away, otherwise keep it in the list. The longest strings are kept at the end. Once you have a string of length n, return it.

Python 2/3, 236 bytes

236 byte version:

n=input();r=len;u=[("",[0]*(n//4))]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1;
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

This is fairly fast, for a brute-force-ish method, only taking a few seconds for n=223, but much longer for n>=224.

Python 2/3, 236 bytes

236 byte version:

n=input();r=len;u=[("",[0]*(n//4))]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1;
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

This is fairly fast, for a brute-force-ish method, only taking a few seconds for n=223, but much longer for n>=224.

Explanation: Keep track of a list of string-list pairs (s,u), where the list u is such that u[i] is the current position after following every ith step in the string. For each string in the list, try to add "L" or "R", then change the values in the list that intersect. (i.e. if the resulting string has length 10, add or subtract 1 from positions 1,2,5 and 10, according to the directions you moved). If you exceed 3 or -3 throw the new pair away, otherwise keep it in the list. The longest strings are kept at the end. Once you have a string of length n, return it.

Python 2/3, 253 or 236 bytes

236 byte version:

n=input();r=len;u=[("",[0]*(n//4))]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1;
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

This is fairly fast, only taking a few seconds for n=223. At some point, some golfing I did made it run way too long for n>=224 (note that 224=4*56), though before it would run fine on those numbers. The culprit is the change from dynamically expanding the array of sums to pre-assigning it with the length it will always reach. The following code works fine:

253 byte version:

n=input();r=len;u=[("",[])]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  if-~r(y)//4>=r(s):s+=0,
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

Any explanation of why is appreciated, as 56 seems like an odd number to cause problems. Note that I am using Python 2.7.11 to run my code, which may be affecting this.

Python 2/3, 236 bytes

236 byte version:

n=input();r=len;u=[("",[0]*(n//4))]
while n>r(u[-1][0]):
 y,t=u.pop()
 for c in 0,1:
  s=t[:];u+=(y+"LR"[c],s),
  for i in range(r(s)):
   if-~r(y)//-~i*-~i==-~r(y):s[i]+=2*c-1;
   if abs(s[i])>3:u.pop();break;
print(u[-1][0])

This is fairly fast, for a brute-force-ish method, only taking a few seconds for n=223, but much longer for n>=224.