"Graded free" is stronger than "graded and free"? This topic suggested me the following question: 

If $R$ is a commutative graded ring and $F$ a graded $R$-module which is free, then can we conclude that $F$ has a homogeneous basis (that is, a basis consisting of homogeneous elements)?

In general the answer is negative, and such an example can be found in Nastasescu, Van Oystaeyen, Methods of Graded Rings, page 21. But their example is not quite usual, and 

I'd like to know if however the property holds for positively graded $K$-algebras, for example. 

If not, then maybe someone can provide a generic example when the property holds.
 A: It is not clear to me in the previous answer what assumptions are being made on the $\mathbb{N}$-graded algebra $R$ (particularly, what assumptions are made on $R_0$), but I'll try to be very careful here. Suppose that $R = \bigoplus_{n=0}^\infty R_n$ and that $R$ is connected, meaning that $R_0 = K$.  Let's say that a graded $R$-module $M = \bigoplus_{n \in \mathbb{Z}} M_n$ is bounded below if $M_n = 0$ for $n \ll 0$.
Then I claim that if $R$ is connected, then every bounded-below graded $R$-module that is free has a homogeneous basis. 
This can be deduced, even in a noncommutative setting, in the very informative Section 2 of the paper The structure of AS regular algebras by Minamoto and Mori, especially Lemma 2.6.  (In fact, one can try to drop further assumptions; I suspect that it's enough to assume $R_0$ is a perfect local ring, not necessarily commutative, in which case every bounded-below graded projective left module would probably be free with a homogeneous basis.)  
I'll sketch an argument here; it's essentially the same as the one in the answer already given by user26857.  The key observation is that the graded Nakayama's lemma works for bounded-below modules, of which finitely generated graded modules form a special case.  
Proof of claim: Supposing that $F$ is graded, bounded-below, and free, fix a set of homogeneous elements $\{x_i\} \subseteq F$ whose images $\overline{x_i} \in F/R_+ F$ form a homogeneous basis for the graded vector space.  We get a graded map from a bounded-below graded free module $\phi \colon G = \bigoplus R(-l_i) \to F$, where each $l_i = \deg(x_i)$, sending the $i$th basis element to $x_i$.  Nakayama implies that this is a "minimal projective cover," in the sense that $\phi$ is surjective and $\ker(\phi) \subseteq R_+ G$. Since $F$ is free we have $G \cong F \oplus L$ for the graded submodule $L = \ker(\phi) \subseteq G$.  But now $$L = \ker(\phi) \subseteq R_+ G = R_+(F \oplus L) = R_+ F \oplus R_+ L.$$ Inspecting this direct sum decomposition, we must in fact have $L \subseteq R_+ L$. Nakayama now implies that $\ker(\phi) = L = 0$. Thus $F \cong G$ as graded modules, so that $F$ has a homogeneous basis.  QED
What I would really like to know is what happens if one drops the bounded-below assumption.  Is there an example of a graded module over a connected graded (commutative) ring $R$ that is not bounded-below, and which does not have a homogeneous basis?
A: Only a partial result for now.

If $R$ is an $\mathbb N$-graded $K$-algebra and $F$ is a $\mathbb Z$-graded $R$-module which is free of finite rank, then $F$ is gr-free.

Let $x_1,\dots,x_n$ be a minimal system of homogeneous generators for $F$. Then their images $\bar x_1,\dots,\bar x_n$ in $F/R_+F$ generate the $K$-vector space $F/R_+F$, and therefore we can suppose that $\bar x_1,\dots,\bar x_m$, $m\le n$, form a $K$-basis in $F/R_+F$. Thus we have $F=R_+F+\langle x_1,\dots,x_m\rangle$, and the graded Nakayama's lemma gives us $F=\langle x_1,\dots,x_m\rangle$ hence $m=n$.
This shows that $\dim_KF/R_+F=n$, so $n=\operatorname{rank}F$. But a system of generators consisting of $\operatorname{rank}F$ elements is necessarily a basis, and we are done.
