There are two different effects that dominate the coloration of a flame: blackbody radiation and electronic transitions.
In the first case, the temperature determines the wavelength that will dominate. A familiar example is the notion of "white hot" when heating the same object to two different temperatures. The higher the temperature, the shorter the dominant wavelength becomes, and so the "bluer" the flame becomes. In this process, the key is that the SAME MATERIAL is heated to DIFFERENT TEMPERATURES to change the color.
With electronic transition, the particular element or molecule in question becomes more relevant. When electrons (or more generally, whole molecular/etc systems) are excited to a state of higher energy, they may only shed this energy in fixed amounts, corresponding to fixed amounts of energy. The shorter the wavelength of the emitted light (the bluer it is) the more energy it carries. This explains the existence of spectral lines/bands, and underlies the use of the "burn test" in analytical chemistry. For example, when copper is placed in a flame, it will become colored green. I assume the flame itself becomes green because a very small amount of the copper is actually vaporized and excited, but I've never actually considered why the bulk of the flame becomes colored before.
In the case of electronic transition, the intensity of the flame does not play as strong of a role in general.