If $\varphi: A\to B$ is a (norm-continuous, unital, involutive) homomorphism of $C^*$-algebras, then the image $\varphi(A)$ is closed in $B$ and therefore is a $C^*$-algebra with the $C^*$-norm induced from $B$ (G.Murphy, 3.1.6).
If in addition $A$ is a von Neumann algebra, will $\varphi(A)$ be a von Neumann algebra as well?
No. The "why" depends on your definition of von Neumann algebra.
If you define von Neumann algebra the way it is usually used, it is not an intrinsic notion: both the double commutant and the ultraweak operator topology depend on the environment (i.e. $B(H)$). In that case, take a non-type I factor, say a II$_1$, and take an irreducible representation. Then the image will be dense in some $B(H)$, but it cannot be everything (because in a II$_1$ factor the identity is finite, for instance). So the image is not a von Neumann algebra.
The above shows that the usual definition of von Neumann algebra has its caveats. An intrinsic definition can be given: a C$^*$-algebra that admits a Banach pre-dual. More concretely this means that a von Neumann algebra is a C$^*$-algebra that admits an isometric surjective embedding onto the dual of a Banach space.
So in this second case, if $\varphi$ is injective, its image will be a von Neumann algebra. When $\varphi$ is not injective, it is easy to come up with a counterexample: you can take $A=B(H)$ and $\varphi$ the quotient map onto the Calkin algebra, which is not a von Neumann algebra (in an intrinsic way: it is not C$^*$-isomorphic to any von Neumann algebra).