About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.

In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms ${\bf M}_2 \to {\bf M_2}$, with both sides carrying the natural ${\rm C}^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.

(To get such homomorphisms, let
$$ s_t= \left(\matrix{ 1 & t \\ 0 & 1 } \right)$$ and consider the automorphism of ${\bf M}_2$ given by $x \mapsto s_t^{-1} x s_t$. A simple calculation shows that the norm-1 element
$$p=\left(\matrix{1 & 0 \\ 0 & 0 }\right)$$
satisfies
$$s_t^{-1}ps_t = \left(\matrix{ 1 & t \\ 0 & 0} \right)$$ and the latter matrix has norm $ > |t|$.)

One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=\ell^1({\bf Z}_+)$ with convolution product (a.k.a. the completion of the polynomial ring ${\bf C}[z]$ in the natural $\ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $\delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $a\in A_+$ definess a continuous unital endomorphism of $A_+$ which sends $\delta_1\mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see

MR0241980 (39 #3315) Reviewed
D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980

which provides the example $a= (\delta_0+\delta_1-\delta_2)/\sqrt{5}$ among others.

To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B={\bf M}_2$ or $A=B=\ell^1({\bf Z}_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a ${\rm C}^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.