If an algorithm requires an integer which can contain the number n (e.g. counting the size of an input array), that integer must take on the order of log(n) space (right?).
If this is the only space which scales with n, is the space complexity of the algorithm O(logn)?
Formally, it depends on the model of computation you're using. In the classical random access machine model, with a fixed word size, simply storing the length n of the input indeed requires O(log n) space (and simple arithmetic on such numbers takes O(log n) time). On the other hand, in the transdichotomous model, where the word size is assumed to grow logarithmically with the input size n, it only requires O(1) space (and time). Other models may yield yet other answers.
In practice, most analysis of algorithms assumes that simple arithmetic on moderate-size integers (i.e. proportional to the input length) can be done in constant time and space. A practical reason for this, besides the fact that it simplifies analysis, is that in practice the logarithm of the input length cannot grow very large — even a computer capable of counting up from zero to, say, 2256, much less of reading that many bits of input, is probably forever beyond the means of humankind to build using known physics. Thus, for any conceivable realistic inputs, you can simply assume that a machine with a 256-bit word size can store the length of the input in a single word (and that machines with a smaller word size still need only a small constant number of words).