Is the difference between consecutive prime numbers always an even number?

You are thinking way too hard about this.

First, to the question in the title (but not as asked in the text) no: $3-2=1$.

As asked in the text for odd primes, then the difference between two odd numbers is always an even number: $$2p+1 - (2q+1) = 2(p-q).$$

Think binary. For all odd primes the last digit, in binary, is 1. So, for any two odd primes:

P1 = ...1 P2 = ...1 X = P1 - P2 = (+/-)...0

Where ... represents whatever leading digits there are.

Yes, always. If we get two prime numbers $\gt 2$, then both are odd and difference two odd numbers is always even. This can be simply proven:
Each prime number $\gt 2$ is odd. This can be proven by contradiction: If those number is even, then this number is divisible by $2$ and then it is not a prime number.
Difference two odd numbers is even always. Let us have some two odd numbers $p_1$ and $p_2$. These numbers are odd and according to this we can express these numbers as $p_1=2k+1$ and $p_2=2l+1$, where $k\ge 1$ and $l \ge k$, then $\Delta = p_2 - p_1 = 2l+1-2k-1=2(l-k)$. This number is even and difference two odd, including prime, numbers($\gt 2$) is always even number.