Here is a rough construction of such a decomposition. First, glue another copy of $M$ to $M$ along their joint boundaries to get rid of the boundary, and then reduce your question to $S^3$ by performing surgery on a framed link $K\subset M$ that lives inside a ball $B$ that is disjoint from $L$.
Next, form a handlebody by starting with a tubular neighbourhood of $L$, choosing a diagram for $L$, and adding a small `bridging cylinder' between each overcrossing cylinder and its corresponding undercrossing cylinder ("overcrossing" and "undercrossing" determined by the diagram). Note that this handlebody lies entirely outside $B$. If needed, add further bridging cylinders to make the complement also a handlebody. The handlebody thus constructed has a handle decomposition with only $0$--handles and $1$--handles which we complete to a handle decomposition of $S^3$ that satisfies all three conditions which you listed.
Finally, to obtain a Heegaard splitting of $M$ which induces a handle decomposition of $M$ satisfying your conditions, choose a Heegaard splitting of $M$ so that $B$ is inside one of the handlebodies, and connect-sum your handlebody outside $B$ (the one containing $L$) with the other handlebody inside $M$, adding bridging cylinders as required.