Andrej Bauer rightly points out that in general, it does not make sense to ask whether a particular machine/program/gadget is Turing complete.
However, it does make sense when such a device is programmable: when it can read a specification of what it needs to do, and then do that. In other words: when it implements a model of computation. We might call a programmable device Turing complete when the model it implements is Turing complete.
(Andrej Bauer maintains that we call such devices universal, not Turing complete. Let's just say that I have seen the term Turing completeness being applied to devices and to programs, e.g. Minecraft, Doom, and Vim.)
In this sense, every universal Turing machine is Turing complete (universal): it is programmable, and it implements the model of computation of Turing machines. However, most Turing machines aren't programmable, and most programmable Turing machines do not implement a Turing complete model of computation. So the answer to your question is still: no.
For instance, consider a Turing machine that reads a decimal number, then a hash mark, then an arbitrary string up to the next hash mark, and then outputs Y if the string has as many characters as the decimal number indicates, and N otherwise. We can say that this Turing machine is programmable, with each of its programs being a decimal number and the meaning of each program being to decide whether the given input string has the given number of characters. Clearly, this programming language, if you want to call it that, isn't Turing complete by a long stretch, so the machine isn't, either.