How about this approach:
Let N = number of matches that are left in the box. I take 1 match on the first turn, and from then on I always take a quantity of matches that will leave my opponent with either N modulo 7 = 0 or N modulo 7 = 2 matches. In other words, after each turn I want to leave him with one of the following conditions:
If I take 1 on the first move, that leaves him with 49. No matter whether he takes 1, 3 or 4 I can always take an appropriate number on my next turn to get him either to 42 (which evenly divides by 7) or 44 (which divides by 7 with remainder 2). If I continue with this strategy my opponent will ultimately end up with either 0 (in which case I've already won), or 2, in which case I win because his only option is to take 1, leaving me to take 1 for the win.