[EDIT: This question as written contains a glaring error. It's sort of "not a real question"? But I am resisting the temptation to delete it, because I am going to make it into a real question.]

I tend to pay a lot of attention to when basepoints matter and when they don't. Currently something is bothering me:

Consider the "homotopy category" of unbased $1$-connected (i.e. simply-connected) spaces: that is, start with spaces that have one path component and trivial fundamental group, with unbased maps as the morphisms, and invert the weak homotopy equivalences. Or, if you prefer, start with $CW$ spaces of that kind and pass to homotopy classes. I suppose that this is NOT actually (equivalent to) the homotopy category of any model category, which is why I have used quotation marks above.

[EDIT: As Fernando Muro points out in his answer, the previous sentence is obviously wrong. The model structure described in the following paragraph, with its artificial use of a basepoint, does the job!]

If I do the same thing with basepoints, there's no problem. For example, I can take the usual category of based spaces but declare the weak equivalences to be the maps inducing isomorphisms of $\pi_j$ for all $j\ge 2$ at the basepoint. Then every object is equivalent to a $1$-connected space (in fact, if you use the usual fibrations, then all cofibrant objects are $1$-connected), and the homotopy category is as intended. The same kind of thing works for $k$-connected based spaces for other values of $k$.

When $k=0$ it is easy to see that the "homotopy category" of $k$-connected (i.e. path-connected) unbased spaces is not a homotopy category of a model category. In fact, it lacks coproducts. (Exercise: there is no universal example, up to homotopy, of a path-connected space equipped with two maps from the circle.) This is enough, because the coproduct (colimit of discrete diagram) of cofibrant objects in a model category is always a homotopy colimit.

When $k=-1$ (so that $k$-connected spaces means nonempty spaces) it is even easier to see: The homotopy category of nonempty spaces has no initial object.

[EDIT: I will leave this as it stands even though the answer is, come to think of it, obviously "no".]

QUESTION 1: Is it true that the homotopy category of (unbased) $1$-connected spaces cannot be the homotopy category of a model category?

This is not so obvious. You can't just say something like "a homotopy category must have all limits and colimits" because homotopy limits and colimits are not in fact limits and colimits in a homotopy category in general -- the homotopy category of spaces does not have pullbacks.

(I presume that most of what I said above more or less applies to other popular ways of axiomatizing homotopy theory, such as $\infty$-categories.)

Apart from this technical question, I have a vague philosophical question:

QUESTION 2: If the answer to Question 1 is "yes", what is the right response?

I am thinking of rational homotopy theory, for example. I recall that when Quillen worked out an indirect equivalence between on the one hand $1$-connected spaces and rational equivalences and on the other hand $0$-connected commutative differential graded coalgebras over $\mathbb Q$ he used based spaces and coaugmented coalgebras. It's clear why. But Sullivan's equivalence between rational homotopy theory of simply connected spaces and commutative dgas works without basepoints, right? He doesn't do it in a model category framework and (this is really beside the point) he needs finite type conditions. It seems unnatural to insist on based spaces just for the sake of (model category or other) axioms. On the other hand they are such nice axioms ...