Not necessarily. Consider a Caesar shift cipher on the Roman alphabet of 26 characters. We map the letter to one of the numbers 0-25, say $x$ and add a key value $k\in [0,25]$ compute $y=x+k\mod {26}$ and then map back to the alphabet. If $k$ is chosen uniformly at random then this is perfectly secure. However, if we enlarge the range of $k$ to say $[0,30]$ this is no longer perfectly secure as values $x+0\mod {26}$, $x+1\mod{26}$, $x+2\mod{26}$, $x+3\mod{26}$ and $x+4\mod{26}$ are twice as likely as the other cipher texts. This gives significant information about $x$ and hence the plaintext. For example, if we see the ciphertext "b" corresponding to $y=1$ we have more evidence that $x=23, 24, 25, 0, 1$ than the other values. Bayesian statistics therefore increases our belief that the plaintext letter lies in the set {'x','y','z','a','b'} and decreases our belief that it lies outside this set. We would not be able to make this inference with a perfectly secure cipher.
Typically, to achieve the uniformity required for perfect security, the key space needs to be a multiple of a size of the ciphertext space and the keys selected uniformly at random. However one can achieve perfect security by other means (e.g. in the above scheme, if we select the keys $\{0,1,2,3,4,26,27,28,29,30\}$ with probability 1/52 and other keys with probability 1/26, then the shift cipher is still perfectly secure.
