Pyth - 21 19 18 bytes

I wonder if there's a trick.

l{st#mP+Q^d3s_BMSE

Test Suite.

l                   Length
 {                  Uniquify
  s                 Combine divisor lists
   t#               Filter by if more than one element
     PM             Take prime factorization of each number
       +RQ          Add each num in list to input
          s_BM      Each num in list and its negative (with bifurcate)
              ^R3   Cube each num in list
                 SE Inclusive unary range - [1, 2, 3,... n] to input

Jelly, 16 bytes

ŒRḟ0*3+µÆfFœ-µQL

Takes x and n as command-line arguments, in that order. Try it online!

How it works

ŒRḟ0*3+µÆfFœ-µQL  Main link. Arguments, x, n

ŒR                Range; yield [-x, ..., x].
  ḟ0              Filter out 0.
    *3            Cube each remaining integer.
      +           Add n to all cubes.
       µ          Begin a new, monadic link. Argument: A (list of sums)
        Æf        Factorize each k in A.
          F       Flatten the resulting, nested list.
           œ-     Perform multiset difference with A.
                  If k in A is prime (or 0), Æf returns [k], adding on k too many
                  to the flat list. Multiset difference with A removes exactly one
                  k from the results, thus getting rid of primes (and 0).
                  If k is composite (or 1), it cannot appear in the primes in the
                  flat list, so subtracting it does nothing.
             µ    Begin a new, monadic link. Argument: D (list of prime divisors)
              Q   Unique; deduplicate D.
               L  Compute the length of the result.

05AB1E, 20 19 bytes

Code:

L3mD(«-Dp_Ïvyf`})Úg

Input in the form x, n. Uses CP-1252 encoding.

Try it online!

MATL, 21 bytes

:3^t_h+tZp~)"@Yf!]vun

Input is x, n separated by a newline.

Try it online!

Explanation

:       % take n implicitly. Generate [1,2,...,n]
3^      % raise to 3, element-wise
t_h     % duplicate, negate, concatenate horizontally: [1,2,...,n,-1,2,...-n]
+       % take x implicitly. Add to that array
t       % duplicate
Zp      % array that contains true for primes
~       % logical negate
)       % apply index to keep only non-primes
"       % for each number in that array
  @     %   push that number
  Yf!   %   prime factors, as a column array
]       % end for each
v       % concatenate vertically all factors
u       % remove repeated factors
n       % number of elements of that array. Implicitly display

J, 30 bytes

#@~.@(,@:q:-.0&,)@:+(|#^&3)@i:

This is a dyadic verb, used as follows:

   f =: #@~.@(,@:q:-.0&,)@:+(|#^&3)@i:
   100 f 4
5

Try it here.

Explanation

#@~.@(,@:q:-.0&,)@:+(|#^&3)@i:
                            i:  Range from -x to x
                    (     )@    Apply this verb to the range:
                       ^&3        a) every item cubed
                     |            b) absolute value of every item
                      #           c) every item in a) repeated b) times; this removes 0
                                     and produces some harmless duplication
                   +            Add n to every element of the resulting list
     (          )@:             Apply this verb to the resulting vector:
             0&,                  a) the vector with 0 appended
      ,@:q:                       b) flat list of prime divisors in the vector
                                     (and some extra 0s since we flatten an un-even matrix)
           -.                     c) list b) with elements of a) removed; this gets rid of
                                     the extra 0s and all primes that were in the list
#@~.@                           Remove duplicates and take length

Julia, 112 bytes

f(n,x)=endof(∪(foldl(vcat,map(k->[keys(factor(k))...],filter(i->!isprime(i)&&i>0,[n+z^3for z=[-x:-1;1:x]])))))

This is a function that accepts two integers and returns an integer.

Ungolfed:

function f(n, x)
    # Get all cube distance numbers
    cubedist = [n + z^3 for z = [-x:-1; 1:x]]

    # Filter out the primes and zeros
    noprimes = filter(i -> !isprime(i) && i > 0, cubedist)

    # Factor each remaining number
    factors = map(k -> [keys(factor(k))...], noprimes)

    # Flatten the list of factors
    flat = foldl(vcat, factors)

    # Return the number of unique elements
    return endof(∪(flat))
end

Ruby, 132 120 114 bytes

I'm well aware that this solution still needs a lot of golfing. Any golfing tips are welcome.

require'prime'
->n,x{(-x..x).map{|i|j=n+i**3;j.prime?||(j==n)?[]:j.prime_division.map{|z|z[0]}}.flatten.uniq.size}

Ungolfing:

require 'prime'

def ravenity(n, x)
  z = []
  (-x..x).each do |i|
    j = n + i**3
    m = j.prime_division
    if j.prime? || j == n
      z << []
    else
      z << m.map{|q| q[0]}
    end
  return z.flatten.uniq.size
end

PARI/GP, 79 bytes

(n,x)->omega(factorback(select(k->!isprime(k),vector(2*x,i,n+(i-(i<=x)-x)^3))))

Here is my original straightforward implementation. The optimized version above combines the two vectors into a single, slightly more complicated vector.

(n,x)->omega(factorback(select(k->!isprime(k),concat(vector(x,i,n-i^3),vector(x,i,n+i^3)))))

Python 3.5, 218 bytes:

lambda r,n:len({z for z in{v for f in{t for u in[[r-q**3,r+q**3]for q in range(1,n+1)]for t in u if any(t%g==0 for g in range(2,t))}for v in range(2,f)if f%v==0}if all(z%g!=0 for g in range(2,z))and z%1==0 and z%z==0})

A nice one-lined lambda function, although it may be a bit long. Since I was using Python, I had to come up with my own way of finding the composites for the first step, and then the prime divisors for the last step, so it was not very easy, and this was the shortest I, by myself. could get it to. Nonetheless, it does what it needs to, and I am proud of it. :) However, any tips for golfing this down a bit more are welcome.

Ruby, 138 bytes

->(n,x){require'prime'
v=((-x..x).to_a-[0]).map{|i|n+i**3}.reject{|e|Prime.prime?(e)}
Prime.each(v[-1]).select{|i|v.any?{|e|e%i==0}}.size}

It was a puny challenge. :-)