The recurrence of the $q-$ binomial coefficients gives $\binom{2n}{m}_q$= $q^m \binom{2n-1}{m}_q+ \binom{2n-1}{m-1}_q$ and therefore $\sum_{m=0}^{2n} (-1)^m \binom{2n}{m}_q =\sum_{m=0}^{2n-1} (-1)^m q^m\binom{2n-1}{m}_q .$ Formula $(1-q^m) \binom{2n-1}{m}_q=(1-q^{2n-1}) \binom{2n-2}{m-1}_q $ implies therefore $ \sum_{m=0}^{2n} (-1)^m \binom{2n}{m}_q =(1-q^{2n-1})\sum_{m=0}^{2n-2} (-1)^{m-1}\binom{2n-2}{m-1}_q .$
Edit
A more elegant proof is the following: Let $r(n)=\sum_{m=0}^n (-1)^m \binom{n}{m}_q.$
Consider the $q-$ exponential function $e(x) = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{(1 - q) \cdots (1 - {q^n})}}} .$
Comparing coefficients gives $\sum\limits_{n = 0}^\infty {\frac{{r(n)}}{{(1 - q) \cdots (1 - {q^n})}}} = e( - x)e(x).$
Now $e(x) = \frac{1}{{(1 - x)(1 - qx)(1 - {q^2}x) \cdots }}$ and therefore $e( - x)e(x) = \frac{1}{{(1 - {x^2})(1 - {q^2}{x^2})(1 - {q^4}{x^2}) \cdots }} = \sum\limits_{}^{} {\frac{{{x^{2n}}}}{{(1 - {q^2})(1 - {q^4}) \cdots (1 - {q^{2n}})}}.} $
Comparing coefficients of the last two series gives
$r(2n) = \frac{{(1 - q)(1 - {q^2}) \cdots (1 - {q^{2n}})}}{{(1 - {q^2})(1 - {q^4}) \cdots (1 - {q^{2n}})}} = (1 - q)(1 - {q^3}) \cdots (1 - {q^{2n - 1}}).$