The answer to your first question is "yes." Here is a particularly simple-minded example.

Suppose $n=0$, so that $\mathfrak{gl}(m,n) = \mathfrak{gl}(m)$. Now if $V$ is a graded representation of $\mathfrak{gl}(m)$, then it must be the case that $V = V_0 \oplus V_1$ for some $\mathfrak{gl}(m)$-submodules $V_0$ and $V_1$ of $V$, i.e., $V_0$ is the even subspace of $V$ and $V_1$ is the odd subspace of $V$, and these must both be $\mathfrak{gl}(m)$-submodules of $V$ because $\mathfrak{gl}(m)$ is a purely even Lie (super)algebra.

I will assume that we are working over a field $k$ of characteristic not dividing $m$. Then $\mathfrak{gl}(m)$ is isomorphic as a Lie algebra to $\mathfrak{sl}(m) \oplus k \cdot I_m$, where $k \cdot I_m$ denotes the one-dimensional abelian Lie subalgebra of $\mathfrak{gl}(m)$ spanned by the identity matrix (and of course $\mathfrak{sl}(m)$ denotes the subalgebra of trace-zero matrices in $\mathfrak{gl}(m)$).

Now let $V$ be a two-dimensional $k$-vector space with basis $\{v_0,v_1\}$. Consider the subspace of $V$ spanned by $v_0$ to be the even subspace of $V$, and the subspace of $V$ spanned by $v_1$ to be the odd subspace of $V$. We can make $V$ into a $\mathfrak{gl}(m)$-module by having $\mathfrak{sl}(m)$ act trivially on $V$, and by having $I_m.v_0 = v_1$ and $I_m . v_1 = 0$. Now $V$ is not a graded representation of $\mathfrak{gl}(m)$, because the vector space decomposition of $V$ into its even and odd subspaces is not a module direct sum decomposition.

Incidentally, this example is perhaps also the canonical example showing that not every finite-dimensional representation of a reductive Lie algebra need be semisimple.

The answer to your first question is "yes." Here is a particularly simple-minded example.

Suppose $n=0$, so that $\mathfrak{gl}(m,n) = \mathfrak{gl}(m)$. Now if $V$ is a graded representation of $\mathfrak{gl}(m)$, then it must be the case that $V = V_0 \oplus V_1$ for some $\mathfrak{gl}(m)$-submodules $V_0$ and $V_1$ of $V$, i.e., $V_0$ is the even subspace of $V$ and $V_1$ is the odd subspace of $V$, and these must both be $\mathfrak{gl}(m)$-submodules of $V$ because $\mathfrak{gl}(m)$ is a purely even Lie (super)algebra.

I will assume that we are working over a field $k$ of characteristic not dividing $m$. Then $\mathfrak{gl}(m)$ is isomorphic as a Lie algebra to $\mathfrak{sl}(m) \oplus k \cdot I_m$, where $k \cdot I_m$ denotes the one-dimensional abelian Lie subalgebra of $\mathfrak{gl}(m)$ spanned by the identity matrix (and of course $\mathfrak{sl}(m)$ denotes the subalgebra of trace-zero matrices in $\mathfrak{gl}(m)$).

Now let $V$ be a two-dimensional $k$-vector space with basis $\{v_0,v_1\}$. Consider the subspace of $V$ spanned by $v_0$ to be the even subspace of $V$, and the subspace of $V$ spanned by $v_1$ to be the odd subspace of $V$. We can make $V$ into a $\mathfrak{gl}(m)$-module by having $\mathfrak{sl}(m)$ act trivially on $V$, and by having $I_m.v_0 = v_1$ and $I_m . v_1 = 0$. Now $V$ is not a graded representation of $\mathfrak{gl}(m)$, because the vector space decomposition of $V$ into its even and odd subspaces is not a module direct sum decomposition.

Incidentally, this example is perhaps also the canonical example showing that not every finite-dimensional representation of a reductive Lie algebra need be semisimple.

EDIT: I overlooked the "positive" assumption in the question, so here is an example more of the type the OP was looking for.

Consider the Lie superalgebra $\mathfrak{gl}(1,1)$. I will assume that we are working over a field of characteristic not equal to $2$. Then $\mathfrak{gl}(1,1)$ admits a basis $\{e,f,I,h\}$ such that $e$ and $f$ are odd, $I$ and $h$ are even, and the only non-trivial commutator relations between these basis vector are $$[e,f]=I,$$ $$[h,e]=2e,$$ $$[h,f]=-2f.$$ Now let $V$ be a two-dimensional $\mathbb{Z}_2$-graded vector space as I defined in my previous example. We can make $V$ into a module for $\mathfrak{gl}(1,1)$ by having $e$, $f$, and $I$ act trivially on $V$, and by having the even element $h$ act by $h.v_0 = v_1$ and $h.v_1 = 0$. As above, this is not a graded representation for $\mathfrak{gl}(1,1)$, since the even subalgebra $\mathfrak{gl}(1,1)_0$ does not act by $\mathbb{Z}_2$-degree preserving endomorphisms.