Yes: an assumption like the one you quote for inaccessibles in L, namely $0^\sharp$. Instead you take a "mouse" i.e. an iterable structure, which has a measure of Mitchell order 1 as the topmost final measure. It is thus a structure N that has two measures with the same critical point, $\kappa$ say. As the second measure gives measure 1 to a set of smaller cardinals below $\kappa$ which are themselves measurable, if one takes a countable such N and iterates the second measure out $On$ many times, then this leaves behind an inner model M, which contains a class of measurable cardinals. Externally this class is cub beneath (and hence contains) all uncountable cardinals.

The mouse N is then a $\sharp$ for the model M; the least such mouse with a measure of order 1 (in the canonical ordering of mice) is indeed countable and is known as ``$0^{sword}$'' in the literature. The assumption of its existence thus ensures the $\sharp$ of a model of the kind you ask for.

(I have just noticed the ``even more'': if one wants higher Mitchell order measures on all uncountable cardinals, then one must assume the existence of mice themselves that have higher Mitchell order and the argument does the same. If one wants a model with uncountable $V-cardinals strong in the model, again this can be effected with the appropriate mouse where the topmost measures concentrates on strong cardinals. And so on and so forth.)