Let $ v = \begin{pmatrix} x \\ y \end{pmatrix} $. Then \[\|v\| = \left\| \begin{pmatrix} x \\ y \end{pmatrix} \right\| = \sqrt{x^2 + y^2},\]and \begin{align*} \left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} v \right\| &= \left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right\| \\ &= \left\| \begin{pmatrix} 2x + 3y \\ -2y \end{pmatrix} \right\| \\ &= \sqrt{(2x + 3y)^2 + (-2y)^2} \\ &= \sqrt{4x^2 + 12xy + 13y^2}, \end{align*}so the given inequality becomes \[\sqrt{4x^2 + 12xy + 13y^2} \le C \sqrt{x^2 + y^2},\]or \[\sqrt{\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}} \le C.\]Thus, we can think of $ C $ as the maximum value of the expression in the left-hand side. Maximizing the expression in the left-hand side is equivalent to maximizing its square, namely \[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}.\]Let $ k $ be a possible value of this expression, which means the equation \[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2} = k\]has a solution in $ x $ and $ y $. We can re-write this equation as \[(4 - k) x^2 + 12xy + (13 - k) y^2 = 0.\]For this quadratic expression to have a solution in $ x $ and $ y $, its discriminant must be nonnegative. In other words, \[12^2 - 4 (4 - k)(13 - k) \ge 0,\]or $ 4k^2 - 68k + 64 \le 0 $. This inequality factors as $ 4(k - 1)(k - 16) \le 0 $. The largest value of $ k $ that satisfies this inequality is 16, so the value of $ C $ we seek is $ \sqrt{16} = \boxed{4} $. Note that equality occurs for \[v = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]