GCD in Gaussian integers.

No, it is not (assuming usual definitions). Indeed, the definition of greatest common divisor is as follows (quoted from Greatest Common Divisor on Wikipedia):

If $R$ is a commutative ring, and $a$ and $b$ are in $R$, then an element $d$ of $R$ is called a common divisor of $a$ and $b$ if it divides both $a$ and $b$ (that is, if there are elements $x$ and $y$ in $R$ such that $dx = a$ and $dy = b$). If $d$ is a common divisor of $a$ and $b$, and every common divisor of $a$ and $b$ divides $d$, then $d$ is called a greatest common divisor of $a$ and $b$.

Thus, in your situation one would say that the elements do not admit a greatest common divisor.

Indeed, there is a specific name for domains where each couple of (non-zero) elements admits a greatest common divisor, these domains are called GCD-domains. Important classes of GCD-domains are UFDs and Bézout domains (i.e., domains where each finitely generated ideal is principal).

(I am a bit puzzled why you mention Gaussian integers, since in fact in the Gaussian integers your situation will never arise, as they are a prinipal ideal domain and thus a GCD-domain.)

In a domain, gcds, if they exist, are unique only up to associateness, i.e. up to a unit multiple. If we wish the gcd to be unique we must choose a unique representative from each associate equivalence class - sometimes called unit normal representatives (esp. in computational algebra).

For example in $\,\Bbb Z$ we may unit normalize gcds by choosing reps $\ge 0$ and in $\,k[x]\,$ we may choose monic reps. You can find literature about algorithms for such by searching on "unit normal gcd".

See this answer for further discussion of such unit normalization.